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Given $x,y,z$ are positive real numbers and $xyz =1$ , prove that $x^2+y^2+z^2+xy+yz+zx \geq 2(\sqrt{x}+\sqrt{y}+\sqrt{z})$

I tried to prove it by using AM-GM-HM inequality on $x,y,z$ to yield the result $(x+y+z)/3 \ge \sqrt{xyz}\ge 3/(xy+yz+xz)$

By further calculations I got the result

$(x^2+y^2+z^2+2(xz+xy+yz) )\ge 9$ [From AM GM]
And
$(xz+yx+zy)\ge3$ [From GM HM]

Subtracting I got $(x^2+y^2+z^2+xy+yz+xz) \ge 6$

But I cannot understand how to bring the RHS of the equation. Any tips ?

user
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Aritra Neogi
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  • Are you sure this is true? If $x=y=z=\frac13$, then $LHS=\frac23$, but $RHS=2\sqrt3$, so $LHS<RHS$. – Sai Mehta Oct 30 '23 at 18:19
  • @SaiMehta: There is a condition $xyz=1$. – Martin R Oct 30 '23 at 18:31
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    Solutions on AoPS: https://artofproblemsolving.com/community/c1068820h2082349p14986582 – Martin R Oct 30 '23 at 18:32
  • I edited the question. This is my first time asking one and I am not so familiar with coding. So please forgive the inaccurate question I posted. – Aritra Neogi Oct 30 '23 at 18:33
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    As a heads up, I generally avoid "subtracting" inequalities, because you have to be very careful with the sign. EG What you did was say that $ A \geq B, C \geq D \Rightarrow A - C \geq B - D$, which is not correct $\quad$ I recommend just "adding" inequalities, namely $ A \geq B , C \geq D \Rightarrow -D \geq -C \Rightarrow A - D \geq B - C$, which is the correct statement. – Calvin Lin Oct 30 '23 at 18:58

1 Answers1

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Using $xyz=1$ and the AM-GM inequality, we get that $$\begin{align}x^2+y^2+z^2+xy+yz+zx&=x^2+y^2+z^2+\frac{xyz}z+\frac{xyz}x+\frac{xyz}y\\&=x^2+y^2+z^2+\frac1z+\frac1x+\frac1y\\&=\left(x^2+\frac1x\right)+\left(y^2+\frac1y\right)+\left(z^2+\frac1z\right)\\&\ge2\sqrt x+2\sqrt y+2\sqrt z\end{align}$$

Sai Mehta
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