If $x,y,z \in[0,1/2]$, with $x+y+z=1$, then prove that: $$\sqrt{1-x^2} + \sqrt{1-y^2} + \sqrt{1-z^2}\geq 4\sqrt{\frac{3-(x^2+y^2+z^2)}{5+x^2+y^2+z^2}}$$

OK so... I've tried to square the expression and note that $1-x^2=a$, $1-y^2=b$ etc, but finally I got a remark about a new "face" of that ineq: $$\frac{2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})}{a+b+c}\geq\frac{8+a+b+c}{8-(a+b+c)}$$

so I appreciate any idea or suggestion..

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    $\begingroup$ @M.Winter Well, I would say this kind of problem only makes sense when the full inequality is presented, thus can barely be summarized by a title. In this sense "Hard inequality" is not that bad as a choice... $\endgroup$ – Cave Johnson Jul 26 '18 at 20:37
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    $\begingroup$ @CaveJohnson The title goes against most of the guidelines. It fits to 20% of the questions on Math.SE. I would not wonder if this title was actually showing a "lazy title" warning while writing the question. Including the inequality in the title is completely appropriate. $\endgroup$ – M. Winter Jul 26 '18 at 20:42
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    $\begingroup$ @M.Winter Guidelines are there to guide us for a good title. Unfortunately, sometimes a good title merely does not exist, or cannot be thought of by our simple mind. At least in my opinion, such inequality is too "big" to fit in a title. This is also the reason that plenty of questions with tag [inequality] combining with [contest-math] have a poor title. $\endgroup$ – Cave Johnson Jul 26 '18 at 20:48
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    $\begingroup$ The ratio (l.h.s)/(r.h.s) is a function symmetric in $x,y,z$ therefore attains its minimum when $x=y=z=1/3$, the corresponding minimum value being $1$. Q.E.D $\endgroup$ – uniquesolution Jul 26 '18 at 21:00
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    $\begingroup$ @uniquesolution: Doesn't one need to do a little more work to establish that it's a global minimum? It's only guaranteed to be a local extremum, isn't it? $\endgroup$ – Brian Tung Jul 26 '18 at 21:26

By Holder $$\left(\sum_{cyc}\sqrt{1-x^2}\right)^2\sum_{cyc}\frac{(3-x)^3}{1-x^2}\geq\left(\sum_{cyc}(3-x)\right)^3=8^3=512.$$ Thus, it's enough to prove that $$\frac{32(5+x^2+y^2+z^2)}{3-x^2-y^2-z^2}\geq\sum_{cyc}\frac{(3-x)^3}{1-x^2}.$$ Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.

Hence, $\frac{32(5+x^2+y^2+z^2)}{3-x^2-y^2-z^2}>0$ and does not depend on $w^3$.

Also, $$\prod_{cyc}(1-x^2)=-x^2y^2z^2+1+\sum_{cyc}(x^2y^2-x^2)=-w^6+A(u,v^2)w^3+B(u,v^2)$$ and $$\sum_{cyc}(3-x)^3(1-y^2)(1-z^2)=\sum_{cyc}(27-27x+9x^2-x^3)(1-y^2-z^2+y^2z^2)=$$ $$=\sum_{cyc}(-x^3y^2z^2+9x^2y^2z^2)+C(u,v^2)w^3+D(u,v^2)=26w^6+C(u,v^2)w^3+D(u,v^2),$$ which says that the inequality $$\frac{32(5+x^2+y^2+z^2)}{3-x^2-y^2-z^2}\geq\sum_{cyc}\frac{(3-x)^3}{1-x^2}$$ is equivalent to $f(w^3)\geq0,$ where $$f(w^3)=-\left(26+\frac{32(5+x^2+y^2+z^2)}{3-x^2-y^2-z^2}\right)w^6+E(u,v^2)w^3+F(u,v^2),$$ which is a concave function.

Here, $A$, $B,$ $C,$ $D,$ $E$ and $F$ they are functions of $u$ and $v^2$ and don't depend on $w^3$.

But the concave function gets a minimal value for an extreme value of $w^3$, which happens in following cases.

  1. One of our variables is equal to $\frac{1}{2}.$

Let $z=\frac{1}{2}.$

Hence, $y=\frac{1}{2}-x$, where $0\leq x\leq\frac{1}{2}.$

We need to prove that $$848x^6-1272x^5-1112x^4+1642x^3-69x^2-184x+39\geq0,$$ which is true by AM-GM: $$848x^6-1272x^5-1112x^4+1642x^3-69x^2-184x+39=$$ $$=39-184x(1-2x)-437x^2(1-2x)^2-106x^3(1-2x)^3=$$ $$=39-92(2x(1-2x))-\frac{437}{4}(2x(1-2x))^2-\frac{53}{4}(2x(1-2x))^3\geq$$ $$\geq39-92\left(\frac{2x+1-2x}{2}\right)^2-\frac{437}{4}\left(\frac{2x+1-2x}{2}\right)^4-\frac{53}{4}\left(\frac{2x+1-2x}{2}\right)^6=$$ $$=39-23-\frac{437}{64}-\frac{53}{256}>0;$$

  1. $w^3=0$.

Let $z=0$ and $y=1-x.$

Hence, $1-x\leq\frac{1}{2},$ which gives $x=\frac{1}{2}$ and we checked it in the previous case.

  1. Two variables are equal.

Let $y=x$.

Thus, $z=1-2x\leq\frac{1}{2},$ which gives $\frac{1}{4}\leq x\leq\frac{1}{2}.$

This substitution gives: $$(3x-1)^2(1-x)(x^3+9x^2+7x-1)\geq0,$$ which is obvious.


  • $\begingroup$ Why someone down voted? $\endgroup$ – Michael Rozenberg Jul 27 '18 at 6:49

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