Let $x$,$y$ and $z$ are positive and $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\leq 3$$

Prove that: $$xy\sqrt{z}+yz\sqrt{x}+zx\sqrt{y}\geq x+y+z$$ The things I have done so far $$3\geq \sum \limits_{cyc}\frac{1}{x}\geq \frac{9}{\sum \limits_{cyc}x}\Rightarrow \sum \limits_{cyc}x\geq 3$$ Then, I tried to use AM-GM and Cauchy -Schwarz but without success.

  • $\begingroup$ let x=y=z=.5 ; xy√z+yz√x+zx√y = .53... < x+y+z = 1.5 $\endgroup$ – batsplatsterson Aug 7 '18 at 11:24
  • $\begingroup$ @batsplatsterson if x=y=z=5 then xy√z+yz√x+zx√y=5.5.√5.3=75√5>3.5=x+y+z $\endgroup$ – Shizumi Aoki Aug 7 '18 at 11:30
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    $\begingroup$ Why does it matter that $a$, $b$ and $c$ are positive if they don't appear anywhere else? $\endgroup$ – celtschk Aug 7 '18 at 11:43
  • $\begingroup$ @celtschk - I assumed that was a typographical error, the OP meant x y z $\endgroup$ – batsplatsterson Aug 7 '18 at 12:02
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    $\begingroup$ @batsplatsterson: In your example, $1/x+1/y+1/z = 6 \not\le 3$. Indeed, if $x=y=z$ then that condition reduces to $x\ge 1$, and the inequality to prove reduces to $x^{5/2}\ge x$, which clearly is true for $x\ge 1$. So if a counterexample exists, it cannot have all three values equal. $\endgroup$ – celtschk Aug 7 '18 at 12:05


After homogenization use $uvw$.

Indeed the condition gives $$\frac{3xyz}{xy+xz+yz}\geq1,$$ which says that it's enough to prove that $$\sum_{cyc}xy\sqrt{z}\geq(x+y+z)\left(\frac{3xyz}{xy+xz+yz}\right)^{\frac{3}{2}}.$$ Now, let $yz=a^2$, $xz=b^2$ and $xy=c^2$, where $a$, $b$ and $c$ are positives.

Thus, we need to prove that $$\sum_{cyc}c^2\sqrt{\frac{ab}{c}}\geq\sum_{cyc}\frac{ab}{c}\left(\frac{3abc}{a^2+b^2+c^2}\right)^{\frac{3}{2}}$$ or $$a+b+c\geq(a^2b^2+a^2c^2+b^2c^2)\left(\frac{3}{a^2+b^2+c^2}\right)^{\frac{3}{2}}.$$ Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Hence, we need to prove that $$3u\geq(9v^4-6uw^3)\left(\frac{3}{9u^2-6v^2}\right)^{\frac{3}{2}},$$ which is $f(w^3)\geq0,$ where $f$ is a linear function.

But a linear function gets a minimal value for the extreme value of the variable,

which happens in the following cases.

  1. $w^3\rightarrow0^+$.

Let $c\rightarrow0^+$ and $b=1$.

Thus, we need to prove that $$(a+1)(a^2+1)^{\frac{3}{2}}\geq3\sqrt3a^2,$$ which is true by AM-GM: $$(a+1)(a^2+1)^{\frac{3}{2}}\geq2\sqrt{a}(2a)^{\frac{3}{2}}=4\sqrt2a^2>3\sqrt3a^2;$$ 2. Two variables they are equal.

Let $b=c=1$.

Thus, we need to prove that $$(a+2)\left(\frac{a^2+2}{3}\right)^{\frac{3}{2}}\geq2a^2+1$$ or $g(a)\geq0,$ where $$g(a)=\ln(a+2)+\frac{3}{2}\ln\frac{a^2+2}{3}-\ln(2a^2+1).$$ But, $$g'(a)=\frac{1}{a+2}+\frac{3a}{a^2+2}-\frac{4a}{2a^2+1}=\frac{2(a-1)(2a^3+4a^2+4a+1)}{(a+2)(a^2+2)(2a^2+1)},$$ which gives $a_{min}=1$ and since $g(1)=0$, we are done!

For the proof of the inequality

$$a+b+c\geq(a^2b^2+a^2c^2+b^2c^2)\left(\frac{3}{a^2+b^2+c^2}\right)^{\frac{3}{2}}$$ we can use another way.

We need to prove that $$(a+b+c)^2(a^2+b^2+c^2)^3\geq27(a^2b^2+a^2c^2+b^2c^2)^2,$$ which is true by AM-GM and Holder.

Indeed, $$(a+b+c)^2(a^2+b^2+c^2)^3=\frac{(a+b+c)^2(a^2+b^2+c^2)^6}{(a^2+b^2+c^2)^3}=$$ $$=\frac{(a+b+c)^2\left(\sum\limits_{cyc}(a^4+2a^2b^2)\right)^3}{(a^2+b^2+c^2)^3}\geq\frac{(a+b+c)^2\left(3\sqrt[3]{(a^4+b^4+c^4)(a^2b^2+a^2c^2+b^2c^2)^2}\right)^3}{(a^2+b^2+c^2)^3}=$$ $$=\frac{27(a+b+c)^2(a^4+b^4+c^4)(a^2b^2+a^2c^2+b^2c^2)^2}{(a^2+b^2+c^2)^3}\geq$$ $$\geq\frac{27(a^2+b^2+c^2)^3(a^2b^2+a^2c^2+b^2c^2)^2}{(a^2+b^2+c^2)^3}=27(a^2b^2+a^2c^2+b^2c^2)^2.$$

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    $\begingroup$ Sorry, I don't know what $uvw$ is? $\endgroup$ – Shizumi Aoki Aug 8 '18 at 1:03
  • $\begingroup$ This is more of a puzzle than a hint... $\endgroup$ – JavaMan Aug 8 '18 at 2:04
  • $\begingroup$ @JavaMan I am ready to explain for you. Are you sure that you want to know how to solve this problem? $\endgroup$ – Michael Rozenberg Aug 8 '18 at 3:03
  • $\begingroup$ @MichaelRozenberg Please help me. I need to know the way to solve this inequality. $\endgroup$ – Shizumi Aoki Aug 8 '18 at 3:27
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    $\begingroup$ @MichaelRozenberg I don't care about the problem or how to solve the problem, but your hint was cryptic at best. $\endgroup$ – JavaMan Aug 8 '18 at 4:08

We need to prove that $ xy\sqrt{z}+yz\sqrt{x}+xz\sqrt{y}\geq x+y+z $ $\Leftrightarrow \frac{xy\sqrt{z}+yz\sqrt{x}+xz\sqrt{y}}{xyz}\geq \frac{x+y+z}{xyz}$ $\Leftrightarrow \frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\geq \frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}$ $\Leftrightarrow \frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}+\frac{2}{\sqrt{x}}+\frac{2}{\sqrt{y}}+\frac{2}{\sqrt{z}}\geq \frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{xz}$ $\Leftrightarrow \frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}+\frac{2}{\sqrt{x}}+\frac{2}{\sqrt{y}}+\frac{2}{\sqrt{z}} \geq (\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^{2}$ We have: $3\geq \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\Rightarrow 3(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq (\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^{2}$ Now, we need to prove: $\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}+\frac{2}{\sqrt{x}}+\frac{2}{\sqrt{y}}+\frac{2}{\sqrt{z}}\geq 3(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) $ Using Cauchy's inequality, we have: $\frac{1}{x^{2}}+\frac{2}{\sqrt{x}}=\frac{1}{x^{2}}+\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{x}}\geq 3\sqrt[3]{\frac{1}{x^{2}}.\frac{1}{\sqrt{x}}.\frac{1}{\sqrt{x}}}=\frac{3}{x}$ Similar, we have: $\frac{1}{y^{2}}+\frac{2}{\sqrt{y}}\geq \frac{3}{y};\frac{1}{z^{2}}+\frac{2}{\sqrt{z}}\geq \frac{3}{z}$ So, $\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}+\frac{2}{\sqrt{x}}+\frac{2}{\sqrt{y}}+\frac{2}{\sqrt{z}}\geq 3(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) $


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