For $x$, $y$, and $z$ positive real numbers, find $\frac{z}{x}$ such that $(x,y,z)$ achieves the maximum value of $$\sqrt{\frac{3x+4y}{6x+5y+4z}} + \sqrt{\frac{y+2z}{6x+5y+4z}} + \sqrt{\frac{2z+3x}{6x+5y+4z}}.$$ I found that the maximum value was $\sqrt3$ from Cauchy's Inequality, but I don't know how to actually achieve the maximum. I know that the terms must be proportional, but I don't know how to go from there. Any help on how to proceed would be greatly appreciated, thanks!
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$\begingroup$ Why don't you post your working, then it would be easier to pinpoint how to get it. $\endgroup$– MacavityCommented Nov 11, 2014 at 17:21
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2 Answers
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Here's how I would go about it: As the expression is homogeneous, we can set $6x+5y+4z=1$ WLOG. Then CS inequality gives: $$(\overline{3x+4y}+\overline{y+2z}+\overline{2z+3x})(1+1+1)\ge \left(\sqrt{3x+4y}+\sqrt{y+2z}+\sqrt{2z+3x} \right)^2$$
So we get $\sqrt{3x+4y}+\sqrt{y+2z}+\sqrt{2z+3x} \le \sqrt3$ and equality is possible iff $\dfrac{3x+4y}1=\dfrac{y+2z}1=\dfrac{2z+3x}1 \implies x:y:z = 1:3:6$
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$\begingroup$ Shouldn't it be $\sqrt{1+1+1}$ instead of $1+1+1$? $\endgroup$ Commented Apr 5, 2019 at 15:47
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$\begingroup$ @YuiToCheng It is correct as written. $\endgroup$– MacavityCommented Apr 6, 2019 at 1:15
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It's $(a+b+c)^2\leq3(a^2+b^2+c^2)$ The equality occurs for $3x+4y=y+2z=2z+3x$.
answered Nov 11, 2014 at 17:32