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I came across this interesting inequality, and was looking for interesting proofs. $x,y,z \geq 0$

$$ 2\sqrt{\frac{x^{2}+y^{2}+z^{2}}{3}}+3\sqrt [3]{xyz}\leq 5\left(\frac{x+y+z}{3}\right) $$

Addendum.

In general, when is

$$ a\sqrt{\frac{x^{2}+y^{2}+z^{2}}{3}}+b\sqrt [3]{xyz}\leq (a+b)\left(\frac{x+y+z}{3}\right) $$

true?

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    $\begingroup$ I'm not sure what is interesting here, nor how to measure the interest found in a proof. $\endgroup$ – Asaf Karagila Jul 26 '11 at 15:41
  • $\begingroup$ I deleted my answer because it was thoroughly invalid. $\endgroup$ – anon Jul 26 '11 at 15:59
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    $\begingroup$ @Asaf, These inequalities are usually abbreviated as $2QM+3GM \leq 5AM$ The interesting bit is that it is true that $GM \leq AM \leq QM$. $\endgroup$ – picakhu Jul 26 '11 at 16:00
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    $\begingroup$ @picakhu: Can you change your title both to something that better reflects the question and doesn't use 'interesting?' It seems obvious to me that you find the question interesting, or you wouldn't ask it. $\endgroup$ – davidlowryduda Jul 26 '11 at 17:19
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    $\begingroup$ I see. I guess you are not familiar with mixing variables/EV then. I'll post a comment on this shortly. $\endgroup$ – Soarer Jul 26 '11 at 17:34
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This is not a direct answer to the question, but it's probably too long for a comment, so I'm leaving it as an answer. (In the comments it seems that OP was not familiar with the technique of mixing variables/smoothing, which was used by Honey_S in the link provided to solve the problem; or the (n-1)-EV theorem, so this answer would be a quick exposition of what they are.)

Mixing variables/smoothing:

In inequalities such as $f(a,b,c) \ge 0$, we seek to prove an inequality of the type $f(a,b,c) \ge f(t,t,c)$. We expect to iterate this inequality so that we can conclude the minimum would be attained when many variables are equal.

Example 1: (AM-GM inequality)

We want to show that if $a,b,c > 0$, then $a+b+c \ge 3(abc)^{1/3}$.

Proof. Consider

$f(a,b,c) = a + b + c - 3(abc)^{1/3}$

Now by 2-variable AM-GM, we see that $f(a,b,c) \ge f(\sqrt{ab}, \sqrt{ab}, c)$. You can then imagine that if we keep on doing such smoothing - i.e. next time replace $(\sqrt{ab}, c)$ with 2-tuple of their geometric mean for example, then in infinite time we reach the case where all three variables are equal, and that $f(a,b,c)$ attains its minimum when $a=b=c$, which is 0.

In general there are many choices of $t$. If there's an initial condition on $a+b+c$, you may want to change $(a,b)$ to $\left(\frac{a+b}{2}, \frac{a+b}{2} \right)$ or sometimes $(0,a+b)$ if you guess that equality case of the inequality involves 0. If there is an initial condition on $a^2+b^2+c^2$, you may change $(a,b)$ to $\left(\sqrt{\frac{a^2+b^2}{2}}, \sqrt{\frac{a^2+b^2}{2}}\right)$ etc.

In the AM-GM example, life is nice because $f(a,b,c) \ge f(t,t,c)$ holds unconditionally. Very often, this is not the case. For example, in Honey_S's solution in the link, after assuming $abc=1$ by homogenity he proved that

$f(a,b,c) \ge f(\sqrt{ab},\sqrt{ab},c)$

for

$ f\left({a,b,c}\right) = 5\left({a+b+c}\right)-2\sqrt{3\left({a^{2}+b^{2}+c^{2}}\right)} $

only when $c = \max (a,b,c)$. However in this case, we are left to show that $f(t,t,c) \ge 0$ under the condition $t^2c = 1$. This is a one-variable inequality easily handled by calculus.

If you want to see more examples of smoothing in action, check this thread and the four links in that post, this and this for example.

(n-1)-EV theorem:

This is a theorem that kills many olympiad inequalities. You can see Vasile Cirtoaje's original article here. Basically what it does is that after some tedious calculus checking, many inequalities actually attain its extremum when (n-1) of the n variables involved in the inequality are equal.(And we can use calculus to check the remaining case) Check out theorem 3 and its corollaries in the link.

If you want to see its applications, see the "Applications" section of Vasc's paper, and if you want some more, see here or maybe a recent post on this forum. I hope this would be enough for you now :)

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    $\begingroup$ Thanks for the two methods. This will be a fun read for me. :) Maybe I can teach some to my friends who are still young enough to tackle olympiads! $\endgroup$ – picakhu Jul 26 '11 at 19:07
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Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.

Hence, we need to prove that $f(w^3)\geq0$, where $f(w^3)=5u-3w-2\sqrt{3u^2-2v^2}$.

We see that $f$ is a decreasing function,

which says that it's enough to prove our inequality for a maximal value of $w^3$,

which happens for equality case of two variables.

Indeed, $x$, $y$ and $z$ are non-negative roots of the equation $$(X-x)(X-y)(X-z)=0$$ or $$X^3-3uX^2+3v^2X-w^3=0$$ or $$w^3=X^3-3uX^2+3v^2X$$ and we see that $w^3$ gets a maximal value,

when a line $Y=w^3$ is a tangent line to the graph of $Y=X^3-3uX^2+3v^2X$,

which happens for equality case of two variables.

Since our inequality is homogeneous, we can assume $x=y=1$ and $z=t^3$.

Thus, it's enough to prove that: $$\frac{5}{3}(t^3+2)-3t\geq2\sqrt{\frac{t^6+2}{3}}$$ or $$(5t^3-9t+10)^2\geq12(t^6+2)$$ or $$(t-1)^2(13t^4+26t^3-51t^2-28t+76)\geq0,$$ which is obvious.

Done!

By the same way we can get a best values of $a$ and $b$.

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