Prove that positive $x,y,z$ satisfy $\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}\ge \sqrt{2x}+\sqrt{2y}+\sqrt{2z}$

Question

Prove that positive $x,y,z$ satisfy $$\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}\ge \sqrt{2x}+\sqrt{2y}+\sqrt{2z}$$

Actually, this is a part of my solution to another problem, which is:

If $a,b,c$ are sides of a triangle, prove that $$\sqrt{a}+\sqrt{b}+\sqrt{c}\ge \sqrt{a+b-c}+\sqrt{a-b+c}+\sqrt{-a+b+c}$$

I substituted $a=x+y$, $b=y+z$, $c=z+x$. It's often called "Ravi substitution".

Here's a similar Math.SE question.

Perhaps not as elementary as the solutions below, but both are also direct applications of Karamata's Inequality applied to the concave fn $\sqrt t$. — Macavity, Apr 11 '14 at 01:51

score 5 · Accepted Answer · answered Apr 10 '14 at 22:29

5

It's easy to show that

$$\sqrt{u+v}\ge{\sqrt u+\sqrt v\over\sqrt2}$$

Thus

$$\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}\ge{\sqrt x+\sqrt y\over\sqrt2}+{\sqrt y+\sqrt z\over\sqrt2}+{\sqrt z+\sqrt x\over\sqrt2}=\sqrt{2x}+\sqrt{2y}+\sqrt{2z}$$

answered Apr 10 '14 at 22:29

Barry Cipra

81,321

(the first one essentially boils down to the AM-GM inequality, after taking the square on both sides) – Clement C. Apr 10 '14 at 22:31
1

@ClementC. More like Cauchy-Schwarz, I'd say, as we have $(1+1)(u+v)\ge (\sqrt{u}+\sqrt{v})^2$ after squaring both sides. – user26486 Apr 10 '14 at 22:33
Works too. I was thinking of $$u+v\geq \frac{u+v+2\sqrt{uv}}{2}$$ itself equivalent to $$\frac{u+v}{2}\geq \sqrt{uv}$$ after reorganizing. – Clement C. Apr 10 '14 at 22:34
1

Or just $(\sqrt u-\sqrt v)^2\ge0$ – Barry Cipra Apr 10 '14 at 22:34

score 1 · Answer 2 · answered Apr 08 '17 at 12:13

1

Let $x\geq y\geq z$.

Hence, $(2x,2y,2z)\succ(x+y,x+z,y+z)$ and since $f(x)=\sqrt{x}$ is a concave function,

we are done by Karamata.

answered Apr 08 '17 at 12:13

Michael Rozenberg

203,855

Karamata's inequality had been mentioned in a comment by Macavity, but you added a bit more information. – user26486 Oct 05 '17 at 16:24
@user26486 I did not see this comment, but just solved this problem. – Michael Rozenberg Oct 05 '17 at 16:30

score 1 · Answer 3 · answered Apr 10 '14 at 22:29

\begin{align} & 2(\sqrt{x+y}+\sqrt{x+z}+\sqrt{y+z}) \geq 2(\sqrt{2x}+\sqrt{2y}+\sqrt{2z}) \\ & \Leftrightarrow (2\sqrt{x+y}-\sqrt{2x}-\sqrt{2y})+(2\sqrt{x+z}-\sqrt{2x}-\sqrt{2z})+(2\sqrt{y+z}-\sqrt{2y}-\sqrt{2z}) \geq 0 \end{align}

Note that

\begin{align} & (2\sqrt{x+y}-\sqrt{2x}-\sqrt{2y}) \geq 0 \\ & \Leftrightarrow 2\sqrt{x+y} \geq \sqrt{2x}+\sqrt{2y} \\ & \Leftrightarrow 4(x+y) \geq (\sqrt{2x}+\sqrt{2y})^2=2(x+y)+4\sqrt{xy} \\ & \Leftrightarrow 2(\sqrt{x}-\sqrt{y})^2 \geq 0 \end{align}

Thus $(2\sqrt{x+y}-\sqrt{2x}-\sqrt{2y}) \geq 0$. Similarly $(2\sqrt{x+z}-\sqrt{2x}-\sqrt{2z}) \geq 0$ and $(2\sqrt{y+z}-\sqrt{2y}-\sqrt{2z}) \geq 0$ so we are done. Equality holds iff $x=y=z$.

Prove that positive $x,y,z$ satisfy $\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}\ge \sqrt{2x}+\sqrt{2y}+\sqrt{2z}$

3 Answers3