# Inequality for contests [closed]

Prove that for real numbers $$x,y,z\in[0;1/2]$$ with $$x+y+z=1 :$$ $$\sqrt{1-x^2} + \sqrt{1-y^2} + \sqrt{1-z^2} \geq 2\sqrt{2}$$

## closed as off-topic by Saad, Paul Frost, Aweygan, Chinnapparaj R, José Carlos SantosOct 22 '18 at 18:46

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, Paul Frost, Aweygan, Chinnapparaj R, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.

• Is this for an ongoing contest? The site policy forbids answering questions for an ongoing contest, so when you post a contest problem, it helps if you say what contest it's from. – saulspatz Oct 17 '18 at 16:04
• Try $(x,y,z)=(0.49, 0.26,0.25)$. You will find that this inequality is not true. – Mohammad Zuhair Khan Oct 17 '18 at 16:05
• I'm pretty sure the inequality should be the other way around... Can you check your source? – Isaac Browne Oct 17 '18 at 16:07
• Put that in an edit to the question! – Isaac Browne Oct 17 '18 at 16:09
• I believe that I've proved it with the inequality reversed. Please check the source. – saulspatz Oct 17 '18 at 16:10

## 5 Answers

Note that when $$x=y=z=\frac13$$ we have $$f(x,y,z)=\sqrt{1-x^2}+\sqrt{1-y^2}+\sqrt{1-z^2}=2\sqrt{2}$$ so that to prove the reverse of the inequality in the question it's enough to show that if not all of $$x,y,z$$ are equal, then $$f$$does not assume its greatest value at $$(x,y,z).$$ Suppose then that $$x\neq y.$$ I claim that $$f\left({x+y\over2},{x+y\over2},z\right)>f(x,y,z).$$ We must show that $$2\sqrt{1-\left({x+y\over2}\right)^2}>\sqrt{1-x^2}+\sqrt{1-y^2}$$

This is easy to prove in the usual way, by squaring both sides, collecting like terms, and squaring both sides once again.

Using Cauchy Schwarz, we can write $$(\sqrt{1-x}\sqrt{1+x} + \sqrt{1-y}\sqrt{1+y}+\sqrt{1-z}\sqrt{1+z})^2$$ $$\leq ((1-x) + (1-y) + (1-z)) ((1+x)+(1+y)+(1+z))$$ $$=(3-x-y-z)(3+x+y+z) = (3-1)(3+1)=2\cdot4=8$$ And taking the square root of both sides, we get the reverse of your inequality, which is $$\sqrt{1-x^2}+\sqrt{1-y^2}+\sqrt{1-z^2} \leq 2\sqrt{2}$$

Given that the function $$f(x) = \sqrt{1-x^2}$$ is concave on $$(0, 1/2)$$, \begin{align*} \sqrt{1-x^2}+ \sqrt{1-y^2} + \sqrt{1-z^2} &\le 3 \sqrt{1-\Big(\frac{x+y+z}{3}\Big)^2}\\ &=2\sqrt{2}. \end{align*}

The reverse inequality could be proved by Lagrange multipliers method.

We want to maximize $$f(x,y,z) = \sqrt{1-x^2} + \sqrt{1-y^2} + \sqrt{1-z^2}$$ subject to $$g(x,y,z)=x+y+z=1$$

Taking gradients of both sides, we get, $$x=y=z=1/3$$

Thus we have $$f(1/3,1/3,1/3)= 3(\sqrt 8 /3)= 2\sqrt 2$$

Check a nearby point, such as $$f(.3,.3,.4)=2.82439<2\sqrt 2$$ shows that $$x=y=z=1/3$$ is a maximizer.

The reversed inequality is true.

Indeed, by AM-GM we obtain: $$\sum_{cyc}\sqrt{1-x^2}=\sqrt{\sum_{cyc}(1-x^2+2\sqrt{(1-x^2)(1-y^2)}}\leq$$ $$\leq\sqrt{\sum_{cyc}\left(1-x^2+\frac{1}{2}(1-x+1-y)(1+x+1+y)\right)}=$$ $$=\sqrt{\sum_{cyc}(1-x^2+2-x^2-xy)}\leq\sqrt{\sum_{cyc}(3-x^2-2xy)}=\sqrt{9-1}=2\sqrt2.$$