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Problem: Prove inequality $(x+y+z-2xyz)^2 \le 2\ (1)$ with $x^2+y^2+z^2 = 1 \land x,y,z \in \mathbb R$

I tried expand $LHS$ and have: $$(1)\iff 1 - 2 (xy+yz+xz) + 4 xyz(x+y+z)-(2xyz)^2 \ge 0$$

Denote: $xy = a, yz = b,xz=c \implies (1) \iff1-2\sum a+ 4 \sum ab - 2abc \ge0$

But stuck. Please help me.

Xeing
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2 Answers2

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Expanding the $LHS$ we have

\begin{align}1+ 2(xy+ xz+ yz) - 4xyz(x+y+z) + 4x^2y^2z^2 \leq 3- 4xyz(x+y+z) + 4x^2y^2z^2 \end{align} since $(xy+ xz+ yz) \leq (x^2+y^2+z^2)=1$ by Cauchy-Schwarz inequality. Then \begin{align} 3- 4xyz(x+y+z) + 4x^2y^2z^2 &= 3- 4x^2y^2z^2(1/xy+1/yz+1/xz) + 4x^2y^2z^2\\& \leq 3- 32 x^2y^2z^2 \end{align}

since $\frac{3}{(1/xy+1/yz+1/xz)}\leq \frac{(x^2+y^2+z^2)}{3}\leq \frac{1}{3}$ by harmonic and arithmetic means inequality and the previous estimation.

Paul
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we let $x\le y\le z$,then $$z^2\ge\dfrac{1}{3},2xy\le x^2+y^2\le\dfrac{2}{3}$$

use Cauchy-Schwarz inequality we have $$(x+y+z-2xyz)^2=[(x+y)+z(1-2xy)]^2\le [(x+y)^2+z^2][1+(1-2xy)^2]$$ we only prove $$[(x+y)^2+z^2][1+(1-2xy)^2]\le 2$$

since: $$ [(x+y)^2+z^2][1+(1-2xy)^2]=(1+2xy)[1+(1-2xy)^2]=2+4x^2y^2(2xy-1)\le 2$$ By done!

math110
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