http://www.mathlinks.ro/Forum/viewtopic.php?t=33007 http://www.mathlinks.ro/Forum/viewtopic.php?t=50902 Darij

Let $x,y$ and $z$ are non-negative real numbers satisfying $x+y+z=1$.Prove that \[0\le yz+zx+xy-\lambda xyz\le{{9-\lambda} \over27}.(0<\lambda \le \frac{9}{4})\] See http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=2847133#p2847133

Easily solved using Lagrange multipliers and then little manipulation

For $ a, b, c > 0 $ real numbers, we have: $ (a+b+c)(ab+bc+ca)-2abc \le \frac{7}{27} (a+b+c)^3 \ \ ; $ Greetings!

How do I prove this inequality ? a^n+1 - 1/(a^n+1) > (n+1/na)(a^n - 1/a^n)

$0\le yz+zx+xy-2xyz\le{7\over27}$ since $x+y+z=1$ we can renote $x={a\over(a+b+c)}$ , $y={b\over(a+b+c)}$ , and $z={c\over(a+b+c)}$ with no restrictions over $a$ , $b$ and $c$ easy computations result $0\le 27(a+b+c)(ab+bc+ca) -54abc \le 7(a+b+c)^3$ the left part is easy since $6abc\le ab(a+b)+bc(b+c)+ca(a+c)$ now , for the right part we can just use Schur's inequality $:$ $ab(a+b)+bc(b+c)+ca(a+c) \le a^3+b^3+c^3 +3abc$ and the fact that $ab(a+b)+bc(b+c)+ca(a+b) \le 2(a^3+b^3+c^3)$ PS: i know this is not the best solution

orl wrote: Prove that $$0\le yz+zx+xy-2xyz\le{7\over27}$$, where $x,y$ and $z$ are non-negative real numbers satisfying $x+y+z=1$. MariusStanean: Let $z=\min\{x,y,z\}\Longrightarrow z\le\frac13$ ， then $$ yz+zx+xy-2xyz=xy(1-2z)+z-z^2\ge 0$$$$ yz+zx+xy-2xyz\le (1-2z)\frac{(x+y)^2}{4}+z-z^2=\frac14+\frac{z^2(1-2z)}{4}\le \frac14+\frac14\left(\frac{z+z+1-2z}{3}\right)^3=\frac{7}{27}$$

Lower bound: $xy+zx+zx-2xyz = (xy+yz+zx)(x+y+z)-2xyz = x^2y+x^2z+xy^2+xz^2+yz^2+y^2z+xyz \geq 0$ because $x,y,z$ are nonnegative Upper bound: Since $x+y+z=1$ notice that $yz+zx+xy-2xyz = (\frac{1}{4})(1-2x)(1-2y)(1-2z) + (\frac{1}{4})$ By AM GM $(1-2x)(1-2y)(1-2z) \leq \frac{1}{27}$ Therefore $yz+zx+xy-2xyz \leq (\frac{1}{4})(\frac{1}{27}) + \frac{1}{4} = \frac{7}{27}$. $\blacksquare$

Claim 1: $xy+yz+zx-2xyz \geq 0$ Proof. $x+y+z=1 \implies x,y,z \leq 1 \implies xyz \leq xy \text{ and } xyz \leq yz.$ But now, $xy+yz+xz \geq 2xyz$ and the claim follows. $\square$ Claim 2: $xy+yz+zx-2xyz \leq \frac{7}{27}$ Proof. Note that $xy+yz+xz-2xyz=\frac{(1-2x)(1-2y)(1-2z)}{4} +\frac{1}{4}.$ Since $x+y+z=1$, we have two cases: a) WLOG, $x > \frac{1}{2}$ and $y,z \leq \frac{1}{2} \implies \frac{(1-2x)(1-2y)(1-2z)}{4} +\frac{1}{4} \leq \frac{1}{4} =\frac{7}{28} <\frac{7}{27}$. b) $x,y,z \leq \frac{1}{2} \implies \text{ by AG-GM, } (1-2x)(1-2y)(1-2z) \leq \frac{(3-2(x+y+z))^3}{27}=\frac{1}{27}<\frac{7}{27}.$ And we're done. $\square$

Homogenize; then, this lemma instantly kills the problem. To make this post more than one line long, we present an equally stupid smoothing solution for the maximum. WLOG $z \leq \frac 12$. Observe that \begin{align*} xy+yz+zx-2xyz &= z(1-z)+(1-2z) xy \\ &\leq (1-z)+(1-2z)\left(\frac{1-z}2\right)^2 \\ &\leq \frac 7{27} \end{align*}by taking the derivative with respect to $z$. Yay.

Good! Nice!

I will show the upper bound. Since the inequality is symmetric, W.L.O.G. let $x\leq y\leq z$. Since $x\leq \frac{1}{3} <\frac{1}{2}$ we have $1-2x>0$, then \begin{align*} xy+zx+yz-2xyz &= x(y+z)+yz(1-2x) \\ &\leq x(1-x)+\frac{(1-x)^2}{4}(1-2x) \\ &= \frac{1+x^2(1-2x)}{4} \\ &\leq \frac{1+(\frac{x+x+1-2x}{3})^3}{4} \\ &= \frac{7}{27}. \end{align*}with equality at $y=z$ and $x=1-2x$, i.e. at $x=y=z=\frac{1}{3}$.

Maybe you can do the lower bound like this: Use $t+1/t\ge 2$. If none of $x,y,z$ are zero we can divide out by $xyz$ and note $x+1/x+y+1/y+z+1/z\ge 6$. If $x=0$, say, then the result is obvious.