Since the TSTST is over, I suppose I can post the problem.

Thank you very much.

raxu wrote: 4. Let x, y, and z be real numbers (not necessarily positive) such that $x^4+y^4+z^4+xyz=4$. Show that $x\le2$ and $\sqrt{2-x}\ge\frac{y+z}{2}$. Are you sure you typed it right? Can someone give an example of an equality case for $x=2$?

mathtastic wrote: Can someone give an example of an equality case for $x=2$? It's not necessary because $x\leq\frac{1}{4}\sqrt{1+5\sqrt{41}}$.

Using 2 makes it more pleasing to the eye and makes it easier in proofs.

Notes from graders: We don't know how to state $\sqrt{2-x}$ otherwise, so we included the first statement. It is supposed to worth almost no point.

arqady wrote: mathtastic wrote: Can someone give an example of an equality case for $x=2$? It's not necessary because $x\leq\frac{1}{4}\sqrt{1+5\sqrt{41}}$. Ah, that's what I was getting. It's weird that they'd use a $\leq$ sign, though.

my solution: assume that $x>2$, then $x^4>16>4$ .so there are exactly one among y,z which less than 0.wlog $z<0$. replace z by -z the condition becomes $x^4+y^4+z^4=4+xyz$ and $x,y,z>0$. so $x^4<xyz+4<(yz+2)x$ , $yz>6$.then $y^4+z^4\ge 2y^2z^2>12yz$. so $xyz=x^4+y^4+z^4-4>12yz$ , $x>12$. $x^4<xyz+4<(yz+1)x$ , $yz>12^3-1$ ; $y^4+z^4>2(12^3-1)yz$ , $x>2(12^3-1)$... this can be continued for infinitely many times and we obtain a contradiction. let $x+y+z=p$ , $xy+yz+zx=q$ , $xyz=r$. $\sqrt{2-x}\ge \frac{y+z}{2}\Leftrightarrow x\le 2-\frac{(y+z)^2}{4}\Leftrightarrow f(x)=x^2+(4-2p)x-8+p^2\le 0$ ,and since the last one is a convex function of x we only need to check it when x is maximal or minimal. (notice that this statement is also true when x is replaced by –x;in addition let $f_1(x)=x^2-(4-2p)x-8+p^2$ ) Case 1.$x,y,z\ge 0$ or two of which less than 0.the condition becomes (by adjust the negative variables to its absolute value) $(p^2-2q)^2+4pr-2q^2+r=4$ (1),whose RHS is increasing of p.now we prove that this leads to $p\le 2$. indeed,it only need to prove that when $p=2$ ,(1)RHS$\ge 4$ , which is $12+2q^2+9r\ge 16q$. but this can be easily checked by $q\le \frac{4}{3}$ and $9r\ge 8q-q^3$. now the desired result turns to $f(0)\le 0$ and $f(p)\le 0$ (when $x\ge 0$ in the original condition) and $f_1(0)\le 0$ and $f_1(p)\le 0$ (when $x\le 0$ in the original condition),which is true by $p\le 2$. Case 2.one or three among x,y,z is less than 0. the condition becomes (by adjust the negative variables to its absolute value) $(p^2-2q)^2+4pr-2q^2-r=4$ ,by nearly the same argument we deduce that $p\le 2$ ,so we have $f(0)\le 0$ and $f(p)\le 0$ (when $x\ge 0$ in the original condition) and $f_1(0)\le 0$ and $f_1(p)\le 0$ (when $x\le 0$ in the original condition). thus in all cases we are done.

Lagrange Multipliers is much more efficient at solving the problem. Why did they give such a problem? Is there a nice solution with AM-GM or Cauchy-Schwarz?

I got a solution with AM-GM but I am not sure if it is correct. (I have forgotten half of it already)

Err does this work?

Otherwise, we now prove for $x,y,z \ge 0$ ($y,z<0$ is analogous) with $x^4+y^4+z^4+xyz=4$, that $\sqrt{2-x} \ge \frac{y+z}{2}$. WLOG $x\le1, y,z \ge1$. If we fix $x$, it follows that $y+z$ is at its maximum when one value is at its minimum say $y=1$. Then it suffices to show that if $x^4+z^4+xz=3$, i.e. $z$ is a function of $x$, then $\sqrt{2-x} \ge \frac{z+1}{2}$. But $\sqrt{2-x}$ is concave down and decreasing in $[0,1]$, and $\frac{z+1}{2}$ decreases logarithmically as $x$ increases. Clearly, $x=0$ and $x=1$ satisfy the inequality, so we are done.

For the first part, suppose by way of contradiction that $x > 2.$ Then since $x^4 + y^4 + z^4 > 4$, we must have $xyz < 0.$ It follows that exactly one of $y, z$ is negative. WLOG suppose that this variable is $z$, and set $z_1 = -z.$ It follows that $x^4 + y^4 + z_1^4 = 4 + xyz_1.$ Then by AM-GM, we have \[5 + xyz_1 = x^4 + y^4 + z_1^4 + 1 \ge 4xyz_1 \implies \frac{5}{3} \ge xyz_1.\] Hence, \[x^4 + y^4 + z_1^4 = 4 + xyz_1 \le 4 + \frac{5}{3}.\] But this is an obvious contradiction because the LHS exceeds $4 + \tfrac{5}{3}.$

arqady wrote: $x\leq\frac{1}{4}\sqrt{1+5\sqrt{41}}$. $8x^4=32+x^2-(x+4yz)^2-8(y^2-z^2)^2\leq32+x^2\implies x\leq\frac{1}{4}\sqrt{1+5\sqrt{41}}.$

I was just wondering if anyone has a solution which does not use convexity, concavity, or derivatives e.g. A solution using AM-GM or Cauchy? Thanks!

Does this inequality has a condition that the left and right are equal??

raxu wrote: Let $x$, $y$, and $z$ be real numbers (not necessarily positive) such that $x^4+y^4+z^4+xyz=4$. Show that $x\le2$ and $\sqrt{2-x}\ge\frac{y+z}{2}$. Proposed by Alyazeed Basyoni $4=x^4+(y^2-z^2)^2+2y^2z^2+xyz+\frac{x^2}{8}-\frac{x^2}{8}=x^4+(y^2-z^2)^2+2\left(yz+\frac{x}{4}\right)^2-\frac{x^2}{8}\geq x^4-\frac{x^2}{8}$. Thus, $8x^4-x^2-32\leq0$, which gives $x\leq\frac{1}{4}\sqrt{1+5\sqrt{41}}\leq2$. Let $\sqrt{2-x}<\frac{y+z}{2}$ or $2-x<\frac{(y+z)^2}{4}$. Hence, $8<4x+y^2+z^2+2yz\leq x^4+3+\frac{y^4+1}{2}+\frac{z^4+1}{2}+2yz=4-xyz-y^4-z^4+4+\frac{y^4+z^4}{2}+2yz=$ $=8+(2-x)yz-\frac{y^4+z^4}{2}$, which gives $(2-x)yz>\frac{y^4+z^4}{2}$ and since $2-x>0$, we obtain $2-x>\frac{y^4+z^4}{2yz}$. Thus, $\frac{(y+z)^2}{4}>\frac{y^4+z^4}{2yz}$ or $(y-z)^2(2y^2+3yz+2z^2)<0$, which is contradiction. Done!

We will prove that $x \le \frac{24}{13}$, which implies $x \le 2$. Let $a = \frac{24}{13}$ for convenience. For the sake of contradiction, assume that $x > a$. Note that \[x^4 > a^4 > \left(\frac{3}{2}\right)^4 = \frac{81}{16} > 4, \]so \[x^4 + y^4 + z^4 \ge x^4 > 4. \]This means that we must have $xyz < 0$, so $yz < 0$. We will assume WLOG that $y > 0$ and $z < 0$. Furthermore, we will let $c = z$, so $c > 0$ and \[x^4 + y^4 + c^4 = xyc + 4. \]We will rewrite this relation as \[x(x^3-yc) + (y^4+c^4-4) = 0.\] Now we note that by AM-GM, $x^4 + y^4 + c^4 + 1 \ge 4xyc$, so $5+xyc \ge 4xyc$ and $xyc \le \frac{5}{3}$, which implies \[ yc < \frac{5}{3a}. \]Now $a^4 > \frac{5}{3}$, so \[x^3 > a^3 > \frac{5}{3a} > yc. \]Thus $x^3 - yc > a^3 - yc > 0$. This means that \[x(x^3 - yc) > a(a^3 - yc), \]so \[a(a^3-yc) + (y^4+c^4-4) < 0. \]Since $y^4 + c^4 \ge 2(yc)^2$, we must have \[ 2(yc)^2 - a(yc) + (a^4 - 4) < 0. \]However, I claim that \[ 2X^2 - aX + (a^4 - 4) > 0 \]for all real numbers $X$, which yields our desired contradiction. Indeed, note that the discriminant is \[ \Delta = a^2 - 8(a^4 - 4) < 4 - 8(a^4 - 4) = 8(4 + \frac{1}{2} - a^4). \]From here, it suffices to note that \[a^4 > \left(\frac{3}{2}\right)^4 = \frac{81}{16} > 5 > 4 + \frac{1}{2}, \]implying that $\Delta < 0$, as desired. Now we move on to the second part of the problem. Let $s = y+z$ and $p = yz$ for convenience. (Note that $p \le \frac{s^2}{4}$ always holds.) For the sake of contradiction, we will assume that \[ \sqrt{2-x} < \frac{s}{2}, \]or equivalently $s > 0$ and \[s^2 > 8 - 4x. \]Our given condition, $x^4 + y^4 + z^4 + xyz = 4$, easily becomes \[ 2p^2 + (x-4s^2)p + s^4 = 4-x^4. \]Letting $L$ denote the LHS, we see that \[\frac{\partial L}{\partial p} = 4p + x - 4s^2 \le x - 3s^2. \]Note that $s^2 > 8-4x$, so \[3s^2 > 24 - 12x \ge x \]by our above work and \[ \frac{\partial L}{\partial p} < 0. \]This means that, treating $L$ as a function of $p$, \[ 4 - x^4 = L(p) \ge L(\frac{s^2}{4})\]and thus \[ L(\frac{s^2}{4}) \le 4-x^4. \]This expands to \[ 2(\frac{s^2}{4})^2 + x (\frac{s^2}{4}) \le 4- x^4. \]Letting $u = \frac{s^2}{4}$, we see that $u > 2-x$ and that \[2u^2 + xu \le 4-x^4. \]Letting $M$ denote the LHS, we see that \[\frac{\partial M}{\partial u} = 4u + x > 8 - 3x > 0. \]Thus, treating $M$ as a function of $u$, \[ 4 - x^4 \ge M(u) > M(2-x) \]and so \[ 4 - x^4 > M(2-x). \]Expanding yields \[x^4 + x^2 - 6x + 4 < 0, \]which contradicts \[x^4 + x^2 + 1 + 1 + 1 + 1 \ge 6|x| \ge 6x \]by AM-GM. $\blacksquare$

AlgebraFC wrote: \[\frac{y}{z}+\frac{z}{y}+2-2\left(\frac{y^2}{z^2}+\frac{z^2}{y^2}\right)>0 \implies 2k^2-k-6<0\implies (2k+3)(k-2)<0,\]which is false because $\vert k\vert \geq 2$ It should be \[\frac{y}{z}+\frac{z}{y}+2-2\left(\frac{y^2}{z^2}+\frac{z^2}{y^2}\right)= 2k^2-k+6<0 \]which is not true for all $k$.

Here is a solution from the Thailand TST group's live solve. First part: We will show a slightly stronger bound, which will pay off in the second part. Claim: $x<1.8$ Proof: Eliminate $y$ and $z$ by the following bound. \begin{align*} x^4+y^4+z^4+xyz &\geq x^4+2y^2z^2+xyz \\ &= x^4 + 2\left(yz+\frac x4\right)^2 - \frac{x^2}{8}\\ &\geq x^4-\frac{x^2}{8} \end{align*}Thus $8x^4-x^2\leq 32$, which is equivalent to $|x|\leq\frac{\sqrt{1+5\sqrt{41}}}{4}<1.8$. $\blacksquare$ Second part: Assume for the contradiction that $(y+z)^2>8-4x$. We claim the following bound. Claim: $y^4+z^4+xyz\geq \dfrac{(y+z)^4+2x(y+z)^2}{8}$ Proof: Multiply both sides by $8$ and note the following identity. \begin{align*}8(y^4+z^4)-(y+z)^4 &= (y-z)^2(7y^2+10yz+7z^2) \\ &= 6(y+z)^2(y-z)^2 + (y-z)^4 \\ &\geq (48-24x)(y-z)^2\end{align*}Thus we have $$8(y^4+z^4+xyz)\geq (48-24x)(y-z)^2 - 2x(y-z)^2 = (48-26x)(y-z)^2\geq 0$$as claimed. $\blacksquare$ Now it's easy to finish. Keeping in mind that $(y+z)^2+x > 8-3x\geq 0$, we bound \begin{align*} x^4+y^4+z^4+xyz &\geq x^4+\dfrac{(y+z)^4+2x(y+z)^2}{8} \\ &= x^4+\dfrac{((y+z)^2+x)^2-x^2}{8} \\ &> x^4+\dfrac{(8-3x)^2-x^2}{8} \\ &= x^4+x^2-6x+8 \end{align*}Thus we have $x^4+x^2-6x+4 <0$, which rearranges to $(x-1)^2(x^2+2x+4)<0$, contradiction.

Solved with nukelauncher and Th3Numb3rThr33. It is easy to verify \(x\le2\): for instance, we have by AM-GM that \[0\ge2y^2z^2+x\cdot yz+\left(x^4-4\right).\]The discriminant (as a polynomial in \(yz\)) must be nonzero, i.e. \[0\le x^2-8\left(x^4-4\right)=32+x^2-8x^4.\]It is easy to verify \(x^2\le4\). Now assume that \(y+z\ge0\) and \(2-x<\left(\frac{y+z}2\right)^2\). Expanding, we have \[8<4x+y^2+z^2+2yz.\]Rewriting using the estimates \(4x\le x^4+3\), \(y^2\le\frac{y^4+1}2\), \(z^2\le\frac{z^4+1}2\), we have \[8<\left(x^4+3\right)+\left(\frac{y^4+z^4}2+1\right)+2yz\implies4<x^4+2yz+\frac{y^4+z^4}2.\] Assume for contradiction \(x^4+y^4+z^4+xyz=4\); substituting this for the left-hand side and rearranging, we have \[\frac{y^4+z^4}2<(2-x)yz.\] However, note that \[(2-x)yz<yz\left(\frac{y+z}2\right)^2=\frac{\left(y^3z+yz^3\right)+2y^2z^2}4\le\frac{y^4+z^4}2\]by AM-GM, contradiction.

Observe that by flipping whether $y,z$ are negative or positive, we can guarantee $y+z\ge 0$ while satisfying the condition nevertheless. To see that $x\le 2$, the argument below suffices. We have that \[2y^2z^2 + xyz \le y^4 + z^4 + xyz = 4-x^4.\]Write $yz = a$. We have $2a^2 + xa + x^4 - 4 \le 0$, so we must have $x^2 \ge 8(x^4 - 4)$ since the discriminant $x^2-8(x^4-4)$ is nonnegative. Taking $x^2 = k$, we see that $k \ge 8k^2 - 32$ and $8k^2 - k-32 \le 0$. Thus, we can bound \[k \le \frac{1 + \sqrt{1 + 4 \cdot 8 \cdot 32}}{16} = \frac{1+5\sqrt{41}}{16} < \frac{1+5 \cdot 7}{16} = \frac{36}{16} = \frac{9}{4},\]so \[|x| < \frac{3}{2}.\]Hence the problem amounts to checking \[2-x\ge (y+z)^2/4,\]which also clearly suffices. Now, we apply the method of Lagrange Multipliers. Let \[f(x): = 2-x-(y+z)^2/4,g(x):=x^4+y^4+z^4+xyz=4.\]Let \[\overline{S}=\{\textbf{x}\in [-3/2,3/2]^3\mid g(\textbf{x})=4\}.\]Clearly $f$ achieves a maximum value on $\overline{S}$, say at $\textbf{x}$. The boundary consists of the points at which one of the cyclic variants of the below holds \[|x|=\frac{\sqrt{1+5\sqrt{41}}}{4}.\]At these points, we get \[0=2a^2\pm \frac{\sqrt{1+5\sqrt{41}}}{4}\cdot a+\frac{1026+10\sqrt{41}}{256}-4=2a^2\pm \frac{\sqrt{1+5\sqrt{41}}}{4}\cdot a+\frac{1+5\sqrt{41}}{128},\]so \[a=\mp \frac{\sqrt{1+5\sqrt{41}}}{16}.\]Since $y^2=z^2$, we have $|y|=|z|$ and so equality holds when \[y=\pm \frac{\sqrt[4]{1+5\sqrt{41}}}{4},z=\pm\frac{\sqrt[4]{1+5\sqrt{41}}}{4},x=\pm\frac{\sqrt{1+5\sqrt{41}}}{4}\]where the signs are chosen appropriately. It is not too hard to manually check that the result holds in these cases. Now we move onto the more glamorous part of this application of Lagrange Multipliers: we have \[\langle -1,-(y+z)/2,-(y+z)/2\rangle = \lambda\langle 4x^3+yz,4y^3+xz,4z^3+xy\rangle.\]Clearly the RHS is not all zero, so we get \[4y^3+xz=4z^3+xy\implies (y-z)(4y^2+4yz+4z^2-x)=0.\]We now have two cases. $y=z$: The given rewrites as $x^4+2y^4+xy^2=4$. Since $-(y+z)/2=-y$, we must have $\lambda = -\frac{1}{4y^2+x}$ so $4x^3+y^2=4y^2+x$. This simplifies to $4x^3-x=3y^2$. Thus $y^2=\frac{4x^3-x}{3}$. Substitute this in for the value of $y^2$ in the equation to yield \[4=x^4+\frac 29 (4x^3-x)^2+\frac 13 x(4x^3-x)=\frac{32}{9}x^6+\frac{5}{9}x^4-\frac 19 x^2.\]Let $x^2=r$ and factor \[0=32r^3+5r^2-r-36=(r-1)(32r^2+37r+36).\]It is clear that $32r^2+37r+36\ge 36$, so $r=1$ implying $x=\pm 1$. Then we either get $y=z=1$ or $y^2=-1$, absurd. In the former case, we have equality. $4y^2+4yz+4z^2=x$: We get \[\lambda = -\frac 12 \frac{y+z}{4y^3+4y^2z+4yz^2+4z^3}=-\frac 18 \cdot \frac{1}{y^2+z^2}.\]Thus \[8(y^2+z^2)=4x^3+yz\implies 2x(y^2+z^2)=x^4+xyz/4.\]The given also tells us that $x^4+y^4+z^4+xyz=4$. Thus, we have \[4=y^4+z^4+3xyz/4+2x(y^2+z^2)=y^4+z^4+(2y^2+2z^2+3yz/4)(4y^2+4yz+4z^2)=\]\[9y^4+11y^3z+19y^2z^2+11yz^3+9z^4.\]Thus $\max(|y|,|z|)\ge \sqrt[4]{4/59}$, and so $x\ge 6\sqrt{4/59}>\frac{\sqrt{1+5\sqrt{41}}}{4}$, absurd. Hence equality holds exactly when $x=y=z=1$, and we are done.

Could anyone verify my solution for the 1st part pls : 1st case : If $x\le 1$ then we finished . 2nd one : If $x\ge 1$ and $xyz\ge 0$ thus $x^2 \le x^4+y^4+z^4+xyz=4$ $\implies$ $x\le 2$ 3rd one : If $x\ge 1$ and $xyz \le 0$ then either $y\le 0$ or $z\le 0$ WLOG $z\le 0$ , we may set $z=-z_0$ such that $z_0 \ge 0$ thus the given condinition can be rewritten as $x^4+y^4+z_0^4=4+xyz_0$ with $y$ and $z_0$ being nonnegative real numbers . Now using AM-GM we can easily prove that $xyz_0\le \frac{5}{3}$ And on the other hand we have that $x^3\le xyz_0 +4 \le \frac{17}{3} \le 8$ wich implies that $x\le 2$

Let $x,y,z$ be real numbers such that $x^4+y^4+z^4+xyz=4$.Show that $$x+y+z\leq 3$$$$x+yz\leq 2$$$$x+y+z\leq2+\sqrt{(2-x)(2-y)(2-z)}$$$$\frac{x+y+z}{xy+yz+zx}\le 1+\frac{5}{247}\left( (x-y)^2+(y-z)^2+(z-x)^2\right)$$Let $x,y,z$ be real numbers such that $x^2+y^2+z^2+xyz=4$.Show that $$\frac{x+y+z}{xy+yz+zx}\le 1+\frac{5}{247}\left( (x-y)^2+(y-z)^2+(z-x)^2\right)$$$$x^2+yz\leq4$$$$x^k+yz\leq 2^k$$Where $k\in N^+.$ Let $x,y,z$ be real numbers such that $x^4+y^4+z^4+xyz=4$.Find the maximum of $y+z.$ Let $x,y,z$ be real numbers such that $x^2+y^2+z^2+xyz=4$.Find the maximum of $x $ and $y+z .$

We'll show that $2\ge \left(\dfrac{y+z}2\right)^2+x$ which implies both parts, equivalently $16(x^4+y^4+z^4+xyz)=64\ge((y+z)^2+4x)^2$ which simplifies to $16x^4-16x^2-8xy^2-8xz^2+15y^4+15z^4-4z^3y-4y^3z-6z^2y^2\ge0$, this can be written as $4(z-y)^2(z^2+zy+y^2)+3(z^2-y^2)^2+12(x^2-1)^2+4(x-z^2)^2+4(x-y^2)^2+4(x^4+y^4+z^4+1-4)$ thus it suffices to show that $x^4+y^4+z^4+1\ge4,$ which is true since by AM-GM $x^4+y^4+z^4+1\ge 4|xyz|\ge 4xyz$ and $x^4+y^4+z^4+1+xyz=5\ge 5xyz\implies x^4+y^4+z^4+1\ge 4$

I will actually show that $x<1.5$. Indeed, if $x \geq 1.5$ then $x^4 \geq 4$, so we need exactly one of $y,z$ to be negative; WLOG $z$. Then for positive $y,z':=-z$ the minimum of $y^4+z^4-xyz'$ occurs when $y=z'$, since $y^4+z'^4$ is minimized and $yz'$ is maximized. Then we need $x^4+2y^4-xy^2 \leq 4$ As a quadratic in $y^2$, its minimum occurs when $y^2=x/4$, where it is $x^4+\tfrac{x^4}{2^7}-\tfrac{x^3}{16}$. But in fact it is easy to see that $x^4-\tfrac{x^3}{16}$ is increasing for $x \geq 1.5$ and is greater than $4$ for $x=1.5$, so $x \geq 1.5$ is thus impossible. Furthermore note that swapping the signs of two variables doesn't change the value of $x^4+y^4+z^4+xyz$, so we also have $x>-2$. Thus we will use Lagrange multipliers on $[-1.5,1.5]^3$, with no variables being allowed to lie on the boundary. The relevant constraint function is $g(x,y,z)=x^4+y^4+z^4+xyz$, and we will try to maximize $f(x,y,z)=x+(\tfrac{y+z}{2})^2$ and show that it's at most $2$. Compute $$\Delta f=(1,\tfrac{y+z}{2},\tfrac{y+z}{2}) \text{ and } \Delta g=(4x^3+yz,4y^3+xz,4z^3+xy).$$Any local extrema occur when $\Delta g$ is a real multiple of $\Delta f$. In particular, this means for any local extremum we have $4y^3+xz=4z^3+xy \iff 4(y-z)(y^2+yz+z^2)=x(y-z)$. If $y \neq z$, then $4(y^2+yz+z^2)=x$. Then by AM-GM $yz \leq \tfrac{x}{12}$, and we have $$y^2+yz+z^2=\frac{x}{4} \implies (y+z)^2=\frac{x}{4}+yz \implies x+\left(\frac{y+z}{2}\right)^2=x+\frac{x}{16}+\frac{yz}{4} \leq \left(1+\frac{1}{16}+\frac{1}{12}\right)x,$$but since $x<1.5$ this is at most $2$. Thus $y=z$, which then means we must also have $$y(4x^3+y^2)=4y^3+xy \implies 4x^3-x=3y^2 \implies x+\left(\frac{y+z}{2}\right)^2=x+y^2=\frac{4}{3}x^3+\frac{2}{3}x.$$Thus it suffices to show that $x \leq 1$, which is clearly equivalent to $y \leq 1$. But if both were greater than $1$, then $x^4+2y^4+xy^2>4$, so we're done. $\blacksquare$