stronger Let $x, y,z$ be positive real numbers. Prove that $\sqrt {{x}^{3} \left( y+z \right) }+\sqrt {{y}^{3} \left( z+x \right) }+\sqrt {{z}^{3} \left( x+y \right) }\geq \sqrt {2} \left( xy+xz+yz \right) $

xzlbq wrote: stronger Let $x, y,z$ be positive real numbers. Prove that $\sqrt {{x}^{3} \left( y+z \right) }+\sqrt {{y}^{3} \left( z+x \right) }+\sqrt {{z}^{3} \left( x+y \right) }\geq \sqrt {2} \left( xy+xz+yz \right) $ By Holder,need to prove \[{\frac { \left( {\it \sum} \left( x \right) \right) ^{3}}{{\it \sum} \left( \left( y+z \right) ^{-1} \right) }}\geq 2\, \left( xy+xz+yz \right) ^{2}\] But \[{\frac { \left( x+y+z \right) ^{3}}{ \left( y+z \right) ^{-1}+ \left( z+x \right) ^{-1}+ \left( x+y \right) ^{-1}}}\geq \frac{9}{4}\,{\frac { \left( y+z \right) \left( z+x \right) \left( x+y \right) \left( xy+xz+yz \right) }{x+y+z}}\] <=> \[\sum{(y-z)^2}\geq 0\] \[{\frac { \left( y+z \right) \left( z+x \right) \left( x+y \right) }{ \left( x+y+z \right) \left( xy+xz+yz \right) }}\geq {\frac {8}{9}}\] Done!

$\sum_{cyc} \sqrt{a^2 \cdot a(b+c)} \ge \frac{2}{\frac{1}{a^2}+\frac{1}{a(b+c)}} = \frac{2a^2(b+c)}{a+b+c} $ by $G-H$ now the inequality becomes : $a^2(b+c) \ge \frac{2}{3} \cdot (a+b+c)(ab+bc+ca)$ or after a full expansion $\sum_{cyc} a(b^2+c^2)\ge 6abc$ which is just $A-G$ , $b^2+c^2 \ge 2bc..$ with equality when $a=b=c$

Dear ,you see this: 3-var: $x,y,z>0$,prove that \[{\frac { \left( x+y+z \right) ^{3}}{ \left( xy+xz+yz \right) ^{2}}}\geq \frac{9}{ 4}\,{\frac { \left( y+z \right) \left( z+x \right) \left( x+y \right) \left( \left( y+z \right) ^{-1}+ \left( z+x \right) ^{-1}+ \left( x+y \right) ^{-1} \right) }{ \left( x+y+z \right) \left( xy+x z+yz \right) }}\] Note: \[{\frac { \left( y+z \right) \left( z+x \right) \left( x+y \right) }{ \left( x+y+z \right) \left( xy+xz+yz \right) }}\geq {\frac {8}{9}}\]

4-var Let $a,b,c,d>0$,prove \[{\frac { \left( a+b+c+d \right) ^{5}}{ \left( cda+dab+abc+bcd \right) ^{2}}}\geq \]\[{\frac {256}{27}}\,{\frac { \left( \left( a+b+c \right) ^{-1}+ \left( b+c+d \right) ^{-1}+ \left( a+c+d \right) ^{-1}+ \left( d+a+b \right) ^{-1} \right) \left( a+c+d \right) \left( d+a+b \right) \left( a+b+c \right) \left( b+c+d \right) }{ \left( a+b+c+d \right) \left( cda+dab+abc+bcd \right) }}\]

5-var: Let a,b,c,d,e>0,prove that ${\frac { \left( a+b+c+d+e \right) ^{7}}{ \left( deca+deab+cbde+ebca+ab cd \right) ^{2}}}\geq $ ${\frac {15625}{256}}\,{\frac { \left( \left( a+b+c+ d \right) ^{-1}+ \left( b+c+d+e \right) ^{-1}+ \left( c+d+e+a \right) ^{-1}+ \left( d+e+a+b \right) ^{-1}+ \left( e+a+b+c \right) ^{-1} \right) \left( a+b+c+d \right) \left( b+c+d+e \right) \left( c+d+e +a \right) \left( d+e+a+b \right) \left( e+a+b+c \right) }{ \left( a +b+c+d+e \right) \left( deca+deab+cbde+ebca+abcd \right) }}$

Let $a,b,c,d>0$,prove that \[{\frac { \left( a+b+c+d \right) ^{3}}{ \left( ab+bc+cd+ad+ac+bd \right) ^{2}}}\geq \frac{4}{3} \left( \left( a+b+c \right) ^{-1}+ \left( b+c+ d \right) ^{-1}+ \left( c+a+d \right) ^{-1}+ \left( d+a+b \right) ^{-1 } \right) \]

let $a,b,c,d,e>0$,prove \[{\frac {1}{100}}\,{\frac { \left( a+b+c+d+e \right) ^{5}}{ \left( abc+ bcd+cde+ade+abe \right) ^{2}}}\geq \left( a+b+c+d \right) ^{-1}+ \left( b +c+d+e \right) ^{-1}+ \left( c+d+e+a \right) ^{-1}+ \left( d+e+a+b \right) ^{-1}+ \left( e+a+b+c \right) ^{-1}\]

Welldone!

xzlbq wrote: stronger Let $x, y,z$ be positive real numbers. Prove that $\sqrt {{x}^{3} \left( y+z \right) }+\sqrt {{y}^{3} \left( z+x \right) }+\sqrt {{z}^{3} \left( x+y \right) }\geq \sqrt {2} \left( xy+xz+yz \right) $ more stronger is: $x,y,z>0$,prove that \[\sqrt {{x}^{3} \left( y+z \right) }+\sqrt {{y}^{3} \left( z+x \right) }+\sqrt {{z}^{3} \left( x+y \right) }\geq {\frac {81}{64}}\,{\frac {\sqrt {2} \left( x+y \right) ^{2} \left( y+z \right) ^{2} \left( z+x \right) ^{2}}{ \left( xy+xz+yz \right) \left( x+y+z \right) ^{2}}}\]

this is a hard: $x,y,z>0$,prove \[\left( 2-\sqrt {2} \right) \left( x+y+z \right) ^{2}+ \left( 4\, \sqrt {2}-6 \right) \left( xy+xz+yz \right) \geq \sqrt {{x}^{3} \left( y+ z \right) }+\sqrt {{y}^{3} \left( z+x \right) }+\sqrt {{z}^{3} \left( x+y \right) }\]

$ \sqrt{a^3b+a^3c}+\sqrt{b^3c+b^3a}+\sqrt{c^3a+c^3b} \ge \frac{2}{\frac{1}{a^2}+\frac{1}{a(b+c)}} + \frac{2}{\frac{1}{b^2}+\frac{1}{b(c+a)}} + \frac{2}{\frac{1}{c^2}+\frac{1}{c(a+b)}} = \frac{2a^2(b+c) + 2b^2(c+a) + 2c^2(a+b)}{a+b+c}$. Note that $(a+b+c)(ab+bc+ca) = a^2(b+c) + b^2(c+a) + c^2(a+b) + 3abc$ so we need to prove $2a^2(b+c) + 2b^2(c+a) + 2c^2(a+b) \ge \frac{4(a^2(b+c) + b^2(c+a) + c^2(a+b) + 3abc)}{3}$ or $2a^2(b+c) + 2b^2(c+a) + 2c^2(a+b) \ge 12abc$ which is simple AM-GM.

parmenides51 wrote: Let $a, b,c$ be positive real numbers. Prove that $ \sqrt{a^3b+a^3c}+\sqrt{b^3c+b^3a}+\sqrt{c^3a+c^3b}\ge \frac43 (ab+bc+ca)$ https://artofproblemsolving.com/community/c6h1536643p9272599

Here is 243 IQ solution from me: Since the inequality is homogenious, WLOG $a+b+c=1$. Hence LHS is equal to: $$\sum \sqrt{a^3b+a^3c}=\sum \sqrt{a^3(b+c)}= \sum \sqrt{a^3(1-a)}=\sum \sqrt{ a^3-a^4}$$ Rewrite the inequality as: $$ \sum \sqrt{a^3-a^4}- \frac{4}{3}(ab+bc+ca) \ge 0$$Let $f,g: \mathbb{R} \rightarrow \mathbb{R}$ be the functions $f(a,b,c)=\sum \sqrt{a^3-a^4}- \frac{4}{3}(ab+bc+ca)$ and $g(a,b,c)=a+b+c-1$ The question is equivalent to proving the minimum value of function $f$ subject to $g$ is 0. It is not hard to calculate partial derivatives of these functions: $$f_a=\frac{(3 - 4 a) a}{2 \sqrt{(1-a) a}}+\frac{4}{3}(a-1)$$$$g_a=1$$Noting other partial derivatives are cyclic to $f_a,g_a$. By lagrange multiplers, there is some real number $\lambda$ such that: $$ \lambda \nabla g= \nabla f$$ Since $\nabla g=1$, we have to solve $f_a=f_b=f_c$ to find critical point of function $f$ subject to $g$ $$\frac{(3-4a)a}{2\sqrt{(1-a)a}}+\frac{4}{3}(a-1)=\frac{(3-4b)b}{2\sqrt{(1-b)b}}+\frac{4}{3}(b-1)=\frac{(3-4c)c}{2\sqrt{(1-c)c}}+\frac{4}{3}(c-1)$$One of the obvious solution is $a=b=c$. but graphing the function, we can see there are other solution as $a=b$ and $a\neq c$. It is pretty obvious that there is no solution like $a\neq b\neq c$ by derivative reasons. $a=b=c$ case is obviously satisfying our inequality. We only left with $a=b$ case. Rewrite the inequality as: $$\sqrt{a^4+a^3c}+\sqrt{a^3c+a^4}+\sqrt{c^3a+c^3a} \ge \frac{4}{3}(a^2+2ac)$$$$2\sqrt{a^4+a^3c}+\sqrt{2c^3a}\ge \frac{4}{3}(a^2+2ac)$$By C-S we have $2(a^4+a^3c) \ge (a^2+\sqrt{a^3c})^2$ therefore inequality becomes: $$\frac{2}{\sqrt{2}}(a^2+\sqrt{a^3c})+\sqrt{2c^3a}\ge \frac{4}{3}(a^2+2ac)$$$$\sqrt{2}a^2 + \sqrt{2ac}(a+c) \ge \frac{4}{3}(a^2+2ac)$$Which is trivial by AM-GM hence done.(After this part I realized I could do C-S from start but ok)

Let $M[x,y,z]=\sum a^xb^yc^z$ $$\sum(\sqrt{a^2(ab+ac})\geq \sum(\frac{(\sqrt{2})^3a^2(ab+ac)}{2a^2+ab+ac})\geq \sqrt{2}(\sum(ab))$$$$\Leftrightarrow \sum (2a^3bc(b+c)(2b+a+c)(2c+a+b))\geq (ab+ac+bc)(2a^2+ab+ac)(2b^2+ab+bc)(2c^2+ac+bc)$$$$\Leftrightarrow 14M[3,3,2]+6M[4,2,2]+10M[4,3,1]+2M[5,2,1]\geq 2M[5,2,1]+7M[4,3,1]+8M[4,2,2]+15M[3,3,2]$$$$\Leftrightarrow 3M[4,3,1]\geq M[3,3,2]+2M[4,4,2] \text{, which is true}$$

$$ \begin{aligned} & \sqrt{a^3 b+a^3 c}+\sqrt{b^3 c+b^3 a}+\sqrt{c^3 a+c^3 b} \geqslant \frac{4}{3}(a b+b c+c a) \\ \Leftrightarrow & \sum a(\sqrt{a b+a c}) \geqslant \frac{4}{3}(a b+b c+c a) \\ \Leftrightarrow & \frac{a^2(b+c)}{\sqrt{a b+a c}} \geqslant \frac{4}{3}(a b+b c+c a) \text { and } \text { as } a+(b+c) \geqslant 2 \sqrt{a(b+c)} \\ \Leftrightarrow & \sum \frac{2 a^2(b+c)}{(a+b+c)} \geqslant \frac{4}{3}(a b+b c+c a) \quad(B y A \cdot M \cdot \geqslant G\cdot M) . \\ \Leftrightarrow & \frac{\sum a^2(b+c)}{a+b+c} \geqslant \frac{2}{3}(a b+b c+c a) . \end{aligned} $$After cross multiplying it suffices to prove: $$ \Rightarrow \sum a^2(b+c) \geqslant 6 a b c $$$\begin{array}{rl}\text { which is obvious as }\left(a^2 b+b^2 c+c^2 a\right)+\left(a^2 c+b^2 a+c^2 b\right) \geqslant 3 a b c+3 a b c=6 a b c . \\ B y & A M \geqslant G M .\end{array}$ And, ladies and gentlemen we are done.