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

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

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    $\begingroup$ 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. $\endgroup$ – saulspatz Oct 17 '18 at 16:04
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    $\begingroup$ Try $(x,y,z)=(0.49, 0.26,0.25)$. You will find that this inequality is not true. $\endgroup$ – Mohammad Zuhair Khan Oct 17 '18 at 16:05
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    $\begingroup$ I'm pretty sure the inequality should be the other way around... Can you check your source? $\endgroup$ – Isaac Browne Oct 17 '18 at 16:07
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    $\begingroup$ Put that in an edit to the question! $\endgroup$ – Isaac Browne Oct 17 '18 at 16:09
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    $\begingroup$ I believe that I've proved it with the inequality reversed. Please check the source. $\endgroup$ – saulspatz Oct 17 '18 at 16:10
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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.

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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 $$\sqrt{1-x^2}+\sqrt{1-y^2}+\sqrt{1-z^2} \leq 2\sqrt{2}$$

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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*}

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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.

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The reversed inequality is true.

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

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