Prove that: $\sqrt{x+y} + \sqrt{y+z} + \sqrt{z+x} \leq \sqrt{6(x+y+z)}$

Question

Prove that for nonegative $x,y,z$ we have: $$\sqrt{x+y} + \sqrt{y+z} + \sqrt{z+x} \leq \sqrt{6(x+y+z)}$$

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I prove that using the tangent line method. We may assume that $x+y+z=1$, so you we have to prove $$\sqrt{1-x}+\sqrt{1-y}+\sqrt{1-z}\leq \sqrt{6}$$

A tangent on $f(x)=\sqrt{1-x}$ at $x={1\over 3}$ is $$y=-{\sqrt{6}\over 4}x+{5\sqrt{6}\over 12}$$

So we have, for all $x\in[0,1]$:

$$\sqrt{1-x} \leq -{\sqrt{6}\over 4}x+{5\sqrt{6}\over 12}$$ and we are done...

I wonder if there is elegant method avoiding calculus?

Answer 1

Hint: This is just $$ (a+b+c)^2\leq 3(a^2+b^2+c^2)\iff ab+bc+ca\leq a^2+b^2+c^2 $$ which is proved easily enough. You can start by $a\equiv\sqrt{x+y}$, $b\equiv\ldots$, etc.

Answer 2

We may assume that $x+y+z=1$, so we have to prove $$\sqrt{1-x}+\sqrt{1-y}+\sqrt{1-z}\leq \sqrt{6}$$

By the RMS-AM (root-mean square - arithmetic mean) inequality:

$$ \frac{\sqrt{1-x}+\sqrt{1-y}+\sqrt{1-z}}{3} \le \sqrt{\frac{\left(\sqrt{1-x}\right)^2+\left(\sqrt{1-y}\right)^2+\left(\sqrt{1-z}\right)^2}{3}} = \sqrt{\frac{2}{3}} $$

Answer 3

How about this: $$ (\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x})^2=2(x+y+z)+2\left[\sqrt{(x+y)(y+z)}+\sqrt{(y+z)(z+x)}+\sqrt{(z+x)(x+y)}\right]\leq 2(x+y+z)+(x+2y+z)+(x+y+2z)+(2x+y+z)=6(x+y+z) $$ where one uses the GM-AM inequality. Finally, use $0\leq x\leq y\implies \sqrt{x}\leq\sqrt{y}$.

Answer 4

By your idea:

Let $x+y+z=3$.

Thus, we need to prove that $$\sum_{cyc}\sqrt{3-x}\leq3\sqrt2$$ or $$\sum_{cyc}\left(\sqrt2-\sqrt{3-x}\right)\geq0$$ or $$\sum_{cyc}\frac{x-1}{\sqrt2+\sqrt{3-x}}\geq0$$ or $$\sum_{cyc}\left(\frac{x-1}{\sqrt2+\sqrt{3-x}}-\frac{x-1}{2\sqrt2}\right)\geq0$$ or $$\sum_{cyc}\frac{(x-1)^2}{\left(\sqrt2+\sqrt{3-x}\right)^2}\geq0.$$ Also, by Jensen we obtain: $$\sum_{cyc}\sqrt{x+y}\leq3\sqrt{\frac{\sum\limits_{cyc}(x+y)}{3}}=\sqrt{6(x+y+z)}.$$