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