1
$\begingroup$

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..

$\endgroup$
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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
  • 2
    $\begingroup$ @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
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ @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
1
$\begingroup$

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.

Done!

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.