it is hold for all real number

Darn this is why nobody solved it. Anyway, here is a simple and standard method due to Amol Aggarwal (for all reals $a,b,c$). WLOG $|a|\ge|b|,|c|$, and let $b=xa,c=ya$ for some $x,y\in[-1,1]$. We need to show that \[2-2a(1+x+y)+a^2\left(1+x^2+y^2+\frac{2xy\sqrt{2}}{\sqrt{1+x^2+y^2}}\right)\ge0,\]for all real $a$, which is equivalent to the discriminant in $a$ always being nonpositive, or \[f(x,y)=\frac{4xy\sqrt{2}}{\sqrt{1+x^2+y^2}}-(1+x+y)^2+2(1+x^2+y^2)\ge0\]for all $x,y\in[-1,1]$. This is simple using partials. Assume for the sake of contradiction that there's a critical point with $x\ne y$. Then \begin{align*} f_x &= \frac{4y(y^2+1)\sqrt{2}}{\sqrt{(1+x^2+y^2)^3}}-2(1+x+y)+2(2x) = 0 \\ f_y &= \frac{4x(x^2+1)\sqrt{2}}{\sqrt{(1+x^2+y^2)^3}}-2(1+x+y)+2(2y) = 0, \end{align*}so from the equations $f_x+f_y=0$ and $f_x-f_y=0$, we find that \[x^2+xy+y^2+1=x^3+y^3+x+y\implies(1-x)(1-y)+x^2(1-x)+y^2(1-y)=0,\]i.e. $x=y=1$, a contradiction. Because $f(x,y)$ is symmetric in $x,y$, it suffices to prove the inequality $f(x,y)\ge0$ for $x=y$, $y=-1$, and $y=1$ (which include the critical points and boundary cases), i.e. if $x=y$, \[\frac{4x^2\sqrt{2}}{\sqrt{2x^2+1}}+1\ge4x,\]if $y=-1$, \[x^2+4\ge\frac{4x\sqrt{2}}{x^2+2},\]and if $y=1$, \[x^2+\frac{4x\sqrt{2}}{\sqrt{x^2+2}}\ge4x.\]Equality only holds for the last one at $x=0$. (These are all pretty easy to verify...)

What is TSTST？Could anyone explain it？

Hooksway wrote: What is TSTST？Could anyone explain it？ I think (imo)Team selection test

Mirus wrote: Hooksway wrote: What is TSTST？Could anyone explain it？ I think (imo) That should be ‘Team selection test’。But what is TSTST？

See question 5 here. The short answer is that a small pool of potential team members is chosen over a year in advance through this TSTST. The USAMO will now function as the actual TST.

MellowMelon wrote: See question 5 here. The short answer is that a small pool of potential team members is chosen over a year in advance through this TSTST. The USAMO will now function as the actual TST. Thank you.Do you have the official solution?I want to know the solution of the last problem.

MellowMelon wrote: Let $a, b, c$ be positive real numbers in the interval $[0, 1]$ with $a+b, b+c, c+a \ge 1$. Prove that \[ 1 \le (1-a)^2 + (1-b)^2 + (1-c)^2 + \frac{2\sqrt{2} abc}{\sqrt{a^2+b^2+c^2}}. \] We will prove that the original inequality holds for all nonnegative real numbers $a,\,b,\,c$ satisfying $a+b+c>0$: Since $(1-a)^2+(1-b)^2+(1-c)^2=3-2(a+b+c)+a^2+b^2+c^2,$ the inequality can be written as \[2-2(a+b+c)+a^2+b^2+c^2 +\frac{2\sqrt{2}abc}{\sqrt{a^2+b^2+c^2}} \ge 0.\] Using the AM-GM inequality, we get \[2(a+b+c) \le \frac{(a+b+c)^2}{2}+2\] and \[\begin{aligned} \frac{1}{\sqrt{a^2+b^2+c^2}}& =\frac{\sqrt{2} (a+b+c)}{(a+b+c)\sqrt{2(a^2+b^2+c^2)}} \ge \frac{2\sqrt{2}(a+b+c)}{(a+b+c)^2+2(a^2+b^2+c^2)} \\ &=\frac{2\sqrt{2} (a+b+c)}{3(a^2+b^2+c^2)+2(ab+bc+ca)}.\end{aligned} \] Therefore, it suffices to prove that \[a^2+b^2+c^2-\frac{(a+b+c)^2}{2} +\frac{8abc(a+b+c)}{3(a^2+b^2+c^2)+2(ab+bc+ca)} \ge 0,\] or \[ \frac{16abc(a+b+c)}{3(a^2+b^2+c^2)+2(ab+bc+ca)} \ge 2(ab+bc+ca)-a^2-b^2-c^2.\] If $2(ab+bc+ca) \le a^2+b^2+c^2,$ the above inequality obviously holds. So let us consider now the case $2(ab+bc+ca) \ge a^2+b^2+c^2.$ In this case, we have \[\begin{aligned} 3(a^2+b^2+c^2)+2(ab+bc+ca)& =\frac{8}{3}(a^2+b^2+c^2)+\frac{1}{3}(a^2+b^2+c^2)+2(ab+bc+ca)\\ & \le \frac{8}{3}(a^2+b^2+c^2) +\frac{2}{3} (ab+bc+ca)+2(ab+bc+ca)\\ & =\frac{8}{3}(a^2+b^2+c^2+ab+bc+ca).\end{aligned}\] Thus, we just need to prove the following inequality \[\frac{6abc(a+b+c)}{a^2+b^2+c^2+ab+bc+ca} \ge 2(ab+bc+ca)-a^2-b^2-c^2.\]However, it is obvious because this is the fourth degree Schur's inequality (in fractional form). $\blacksquare$

Let $f(x)=0$ be the function with roots $a,b,c$. Define $g(x)=f(x)-k$, where $k$ is the largest real number so that $g(x)=0$ has three non-negative real roots. (Think of it as sliding the graph of $f(x)=0$ upward along the y-axis as much as possible). The right hand side of the inequality is clearly reduced by this transformation, so it is enough to prove it greater than 1 for the roots of $g(x)=0$. If the smallest root is zero, we are done. If not, then the roots are $tz,z,z$ for some $t \le 1, z\ge 0$. We must show that $(tz-1)^2+2(z-1)^2 +\frac{ 2\sqrt{2}tz^3}{\sqrt{2z^2+t^2z^2}} \ge 1$, or that the quadratic $z^2(t^2+2+\frac{2\sqrt2t}{\sqrt{2+t^2}})-z(2t+4)+2 \ge 0$, for all $z$. This is true if the discriminant is negative, or if $8(t^2+2+\frac{2\sqrt2t}{\sqrt{2+t^2}})\ge 4t^2+16t+16$, which in turn reduces to $t+4\sqrt{\frac{2}{2+t^2}} \ge 4$. For all $0<s<1$, we have the binomial inequality, $(1+s)^{-1/2} \ge 1-s/2$, so $t+4\sqrt{\frac{2}{2+t^2}} \ge t+4(1-\frac{t^2}{4})=4+t^2-t^3 \ge 4$.

Here's a solution with a geometry flavor. Notice if we multiply $a, b, c$ by a factor of $k$, the conditions and the desired inequality change in nice ways. This motivates us to prove the stronger result: Claim. Let $a, b, c, k$ be positive real numbers such that $a, b, c \le k \le a+b, b+c, c+a$. Prove that $$k^2 \le (k - a)^2 + (k - b)^2 + (k - c)^2 + \frac{2\sqrt{2} abc}{\sqrt{a^2 + b^2 + c^2}}.$$Proof. This is a quadratic inequality in $k$, of the form $2k^2 - 2(a+b+c)k + C \ge 0$. It suffices to verify the inequality at the vertex, which occurs at $k = \frac{a + b + c}{2}$. First, notice that $a + b \ge k \ge c$ and similarly for the other variables, so $a, b, c$ form the sides of a (possibly degenerate) triangle. Therefore, we can use the Ravi substitution: $a = x + y, b = x + z, c = y + z$, where $x, y, z \ge 0$. Then $k = x + y + z$. We thus have to show that $$(x + y + z)^2 \le x^2 + y^2 + z^2 + \frac{2\sqrt{2} (x+y)(y+z)(z+x)}{\sqrt{2(x^2 + y^2 + z^2 + xy + yz + zx)}}.$$This is equivalent to $$xy + yz + zx \le \frac{(x + y)(y + z)(z + x)}{\sqrt{x^2 + y^2 + z^2 + xy + yz + zx}}$$or $$(xy + yz + zx)^2 (x^2 + y^2 + z^2 + xy + yz + zx) \le (x + y)^2 (y + z)^2 (z + x)^2.$$However, notice that $$(x + y)(x + z) = x^2 + xy + yz + zx$$and similarly for the other variables. Thus, if we let $xy + yz + zx = B$, then $$(x + y)^2 (y + z)^2 (z + x)^2 = (x^2 + B)(y^2 + B)(z^2 + B) \ge B^3 + (x^2 + y^2 + z^2)B^2$$$$= B^2 (B + x^2 + y^2 + z^2) = (xy + yz + zx)^2 (x^2 + y^2 + z^2 + xy + yz + zx).$$Thus, the required inequality is proved. This implies the stronger result and thus the desired result if we take $k = 1$.

Actually Ravi's substitution suffices, and it's pretty natural in this context (since $a + b \ge c$ is a free condition). Write the original inequality as \[\frac{2\sqrt{2} abc}{\sqrt{a^2+b^2+c^2}} + a^2+b^2+c^2-2(a+b+c)+2 \ge 0\]Suppose $a=y+z$, $b=z+x$, $c=x+y$ for non-negative $x,y,z$. The inequality becomes \[\frac{(x+y)(y+z)(z+x)}{\sqrt{x^2+y^2+z^2+xy+yz+zx}} + (x^2+y^2+z^2+xy+yz+zx) - 2(x+y+z) + 1 \ge 0\]Similar to above, we have \begin{align*} (x+y)^2(y+z)^2(z+x)^2 &= \prod_{\text{cyc}} (x^2+xy+yz+zx) \\ &\ge (xy+yz+zx)^3 + (x^2+y^2+z^2)(xy+yz+zx)^2 \\ &= (xy+yz+zx)^2(x^2+y^2+z^2+xy+yz+zx)\end{align*}so \[\frac{(x+y)(y+z)(z+x)}{\sqrt{x^2+y^2+z^2+xy+yz+zx}} + (x^2+y^2+z^2+xy+yz+zx) - 2(x+y+z) + 1 \ge (x+y+z)^2 - 2(x+y+z) + 1 \ge 0.\]

Here is mine solution i solved this with Kaneki actually i don't know his AoPS id but yeah credits to him and also I m posting this for storage. My Solution: We know that $a+b,b+c,c+a\geq 1$ and $0\leq a,b,c\leq 1$ this satisfies triangle inequality so $a,b,c$ are the sides of the triangle $\bullet$ so lets assume that $a+b,b+c,c+a\geq t$ and $a,b,c\leq t$ where $a,b,c$ are non-negatives. The new inequality becomes like \[t^2\leq (t-a)^2+(t-b)^2+(t-c)^2+\frac{2\sqrt{2} abc}{\sqrt{a^2+b^2+c^2}}\]\[\implies t^2\leq 3t^2+a^2+b^2+c^2-2t(a+b+c)+\frac{2\sqrt{2} abc}{\sqrt{a^2+b^2+c^2}}\]\[\implies 2t^2-2(a+b+c)t+a^2+b^2+c^2+\frac{2\sqrt{2} abc}{\sqrt{a^2+b^2+c^2}}\geq 0\]so here we can see clearly that \[t=\frac{2(a+b+c)}{4}=\frac{(a+b+c)}{2}\] Claim: the inequality is true for $t=\frac{(a+b+c)}{2}$. Proof: The inequality which we have to proof is \[\frac{(a+b+c)^2}{4}\leq \sum_{cyc}\left(\frac{(a+b+c)}{2}-a\right)^2+\frac{2\sqrt{2} abc}{\sqrt{a^2+b^2+c^2}}\]Lets substitute $x=\frac12(b+c-a),y=\frac12(c+a-b),z=\frac12(a+b-c)$ so we get the inequality as \[\implies (x+y+z)^2\leq x^2+y^2+z^2+\frac{2(x+y)(y+z)(z+x)}{\sqrt{x^2+y^2+z^2+xy+yz+zx}}\]we can write this as \[x^2+y^2+z^2+xy+yz+zx\leq \left(\frac{(x+y)(y+z)(z+x)}{xy+yz+zx}\right)^2\]so let $k=xy+yz+zx$ so we have \[k^2(x^2+y^2+z^2+k)\leq (x^2+k)(y^2+k)(z^2+k)\]which is basically true because \[\implies k^3+k^2(x^2+y^2+z^2)\leq k^3+k^2(x^2+y^2+z^2)+k(......)+(xyz)^2\]so we are done !

My solution is quite similar to @above. Since $b+c\ge 1\ge a$, $c+a \ge 1\ge b$, $a+b\ge 1\ge c$. There exist non-negative real numbers $x,y,z$ such that $a=x+y$, $b=y+z$, and $c=z+x$. Note that, $$\frac{2\sqrt{2}abc}{\sqrt{a^2+b^2+c^2}}=\frac{2(x+y)(y+z)(z+x)}{\sqrt{x^2+y^2+z^2+xy+yz+zx}}$$ Claim. $\frac{(x+y)(y+z)(z+x)}{\sqrt{x^2+y^2+z^2+xy+yz+zx}} \ge xy+yz+zx$ Proof. Since both $\frac{(x+y)(y+z)(z+x)}{\sqrt{x^2+y^2+z^2+xy+yz+zx}}$ and $xy+yz+zx$ are non-negative, it suffices to prove $$\left( \frac{(x+y)(y+z)(z+x)}{\sqrt{x^2+y^2+z^2+xy+yz+zx}} \right)^2 \ge (xy+yz+zx)^2$$Let $xy+yz+zx=k\ge 0$, we have $$\left( \frac{(x+y)(y+z)(z+x)}{\sqrt{x^2+y^2+z^2+xy+yz+zx}} \right)^2=\frac{(x^2+k)(y^2+k)(z^2+k)}{x^2+y^2+z^2+k}=\frac{x^2y^2z^2+k(x^2y^2+y^2z^2+z^2x^2)+k^2(x^2+y^2+z^2)+k^3}{x^2+y^2+z^2+k}\ge\frac{k^2(x^2+y^2+z^2)+k^3}{x^2+y^2+z^2+k}=k^2=(xy+yz+zx)^2$$As desired. Now, we have $$(1-a)^2+(1-b)^2+(1-c)^2+\frac{2\sqrt{2}abc}{\sqrt{a^2+b^2+c^2}}=(1-x-y)^2+(1-y-z)^2+(1-z-x)^2+\frac{2(x+y)(y+z)(z+x)}{\sqrt{x^2+y^2+z^2+xy+yz+zx}}=2(x^2+y^2+z^2)+2(xy+yz+zx)-4(x+y+z)+3+\frac{2(x+y)(y+z)(z+x)}{\sqrt{x^2+y^2+z^2+xy+yz+zx}}$$Using the claim, $$2(x^2+y^2+z^2)+2(xy+yz+zx)-4(x+y+z)+3+\frac{2(x+y)(y+z)(z+x)}{\sqrt{x^2+y^2+z^2+xy+yz+zx}}\ge 2(x^2+y^2+z^2)+2(xy+yz+zx)-4(x+y+z)+3+2(xy+yz+zx)=2(x^2+y^2+z^2)+4(xy+yz+zx)-4(x+y+z)+3=2(x+y+z-1)^2+1\ge 1$$As desired.