crazyfehmy wrote: Show that \[ \frac{(b+c)(a^4-b^2c^2)}{ab+2bc+ca}+\frac{(c+a)(b^4-c^2a^2)}{bc+2ca+ab}+\frac{(a+b)(c^4-a^2b^2)}{ca+2ab+bc} \geq 0 \] for all positive real numbers $a, \: b , \: c.$ $\sum_{cyc}\frac{(b+c)(a^4-b^2c^2)}{ab+2bc+ca}\geq0\Leftrightarrow$ $\Leftrightarrow\sum_{cyc}\frac{(b+c)(a^2+bc)\left((a-b)(a+c)-(c-a)(a+b)\right)}{ab+2bc+ca}\geq0\Leftrightarrow$ $\Leftrightarrow\sum_{cyc}(a-b)(a+c)(b+c)\left(\frac{a^2+bc}{ab+ac+2bc}-\frac{b^2+ac}{ab+2ac+bc}\right)\geq0\Leftrightarrow$ $\Leftrightarrow\sum_{cyc}\frac{(a-b)^2(a+c)(b+c)(ab(a+b)+2c(a^2+ab+b^2)-c^2(a+b))}{(ab+ac+2bc)(ab+2ac+bc)}\geq0$. But $a^2+ab+b^2\geq\frac{3}{4}(a+b)^2$. Hence, it remains to prove that $\sum_{cyc}\frac{(a-b)^2(2ab+3c(a+b)-2c^2)}{(ab+ac+2bc)(ab+2ac+bc)}\geq0$. Let $a\geq b\geq c$ and $S_c=(2ab+3ac+3bc-2c^2)(2ab+ac+bc)$. Hence, $S_c\geq0$, $S_b\geq0$ and $a^2S_b+b^2S_a=$ $=a^2(2ac+3ab+3bc-2b^2)(2ac+ab+bc)+b^2(2bc+3ab+3ac-2a^2)(2bc+ab+ac)\geq$ $\geq3ab(ab+ac+bc)(a^2+b^2)-6a^2b^2(ab+ac+bc)=3ab(a-b)^2(ab+ac+bc)\geq0$. Id est, $\sum_{cyc}(a-b)^2S_c\geq(a-c)^2S_b+(b-c)^2S_a\geq$ $\geq\frac{a^2}{b^2}(b-c)^2S_b+(b-c)^2S_a=\frac{(b-c)^2}{b^2}(a^2S_b+b^2S_a)\geq0$. Done!

Here is the official solution. Lemma: $\frac{(b+c)(a^4-b^2c^2)}{ab+2bc+ca} \geq \frac{a^3+abc-b^2c-bc^2}{2}$ Proof: The last one is equivalent to $f(a,b,c)=(b+c)a^4-2bca^3-bc(b+c)a^2+abc(b^2+c^2) \geq 0$ $\frac{f(a,b,c)}{a} \geq \frac{4bc}{b+c}a^3-2bca^2-bc(b+c)a+bc\frac{(b+c)^2}{2}$ $=\frac{bc}{2(b+c)}(2a+b+c)(2a-b-c)^2 \geq 0$ By using this lemma $\sum_{cyc} \frac{(b+c)(a^4-b^2c^2)}{ab+2bc+ca} \geq \frac{a^3+b^3+c^3+3abc-a^2b-ab^2-b^2c-bc^2-c^2a-ca^2}{2}$ Finally, by Schur's inequality $a^3+b^3+c^3+3abc-a^2b-ab^2-b^2c-bc^2-c^2a-ca^2 \geq 0$ and we are done.

crazyfehmy wrote: Show that \[ \frac{(b+c)(a^4-b^2c^2)}{ab+2bc+ca}+\frac{(c+a)(b^4-c^2a^2)}{bc+2ca+ab}+\frac{(a+b)(c^4-a^2b^2)}{ca+2ab+bc} \geq 0 \] for all positive real numbers $a, \: b , \: c.$ After substitution $ab=z$, $ac=y$ and $bc=x$ and full expanding we obtain: $\sum_{sym}(2x^6y^3+7x^5y^4+x^4y^4z-x^6y^2z-3x^5y^2z^2-6x^4y^3z^2)\geq0$, which is obviously true.

crazyfehmy please post solution too Turkey NMO 2009 Question 3

can we make \[xy+xz+yz=1 \] normalization for this inequality? \[f(a,b,c)=S\geq 0 \Leftrightarrow f(ta,tb,tc)=3S\geq 0 (t\geq 0)\] is it enough?

muratsayin wrote: can we make \[xy+xz+yz=1 \] normalization for this inequality? \[f(a,b,c)=S\geq 0 \Leftrightarrow f(ta,tb,tc)=3S\geq 0 (t\geq 0)\] is it enough? The inequality is homogeneous so we can take $ab+bc+ca=1$ but this is not a solution.

crazyfehmy wrote: muratsayin wrote: can we make \[xy+xz+yz=1 \] normalization for this inequality? \[f(a,b,c)=S\geq 0 \Leftrightarrow f(ta,tb,tc)=3S\geq 0 (t\geq 0)\] is it enough? The inequality is homogeneous so we can take $ab+bc+ca=1$ but this is not a solution. I didn't say it is a solution,my solution with xy+xz+yz=1 normalization: \[\sum_{cyc}\frac{(b+c)(a^{4}-b^{2}c^{2})}{1+bc}\] if we clear denominators, \[[4,1,0]+[5,2,0]+[6,2,1]+[5,1,1]\geq [3,2,0]+[3,3,1]+[4,2,1]+[4,3,2]\] it is obvios by muirhead

I have find an alternative solution,but i want write an article about it.

red3 wrote: I have find an alternative solution,but i want write an article about it. what is your alternative solution?write please

I will not post it resently because it seems to be an not known method.But the method is not so strong since i haven't find a proper way to use it . This inequality is obvious from my way.

Does $uvw$ work here?

yunustuncbilek wrote: Does $uvw$ work here? uvw works ! But the proof is complicated Let $u=a+b+c$ ,$v=ab+bc+ca$, $w=abc$ then \[ \frac{(b+c)(a^{4}-b^{2}c^{2})}{ab+2bc+ca}+\frac{(c+a)(b^{4}-c^{2}a^{2})}{bc+2ca+ab}+\frac{(a+b)(c^{4}-a^{2}b^{2})}{ca+2ab+bc}\geq 0 \] Clear the fraction, it becomes \[f(u,v,w)=-7v^4u+2v^3u^3-6v^2u^2+wvu^4+12w^2vu+15wv^3-w^2u^3 \geq 0 \] since the inequality is homogeneous, let assume $u=3$ and take derivative w.r.t w \[f'(u,v,w)=15v^3-54w+81v-54v^2+72wv \] If $v\geq 0.75$ then $f'(u,v,w)=(15v^3+81v-54v^2)+(72wv-54w) \geq 0$ If $v <0.75$ Note $f'(u,v,w)$ is a linear function in w. Also $\frac{2}{9}v-2-\sqrt{27v^2-81v+81-3v^3} \leq w \leq \frac{2}{9}v-2+\sqrt{27v^2-81v+81-3v^3}$ Plug in the boundary for $w$ and the condition $v<0.75$, either way you find that $f'(u,v,w) \geq 0$ So $f(u,v,w)$ is a monotonic increase w.r.t $w$ and we can check $a=b$ or $c=0$

wlog \[a\geq b\geq c\] \[(b+c)(a^4-bc)-(a+c)(b^4-ac)\=\ab(a^3-b^3)+c(a^4-b^4)+c(a^2-b^2)+c^{2}(a-b)\] hence if \[a\geq b\geq c\] then \[(b+c)(a^4-bc)\geq (a+c)(b^4-ac)\geq (a+b)(c^4-ab)\] also \[ab+2bc+ca\leq ab+bc+2ca\leq 2ab+bc+ca\] By chebyshev inequality we obtain \[3\sum\frac{(b+c)(a^4-bc)}{ab+2bc+ca}\geq(\sum a^{4}b-\sum a^{3}b^{2})(\sum\frac{1}{ab+2bc+ca})\geq 0\] QED

I think there is a typo in the above solution By solution runs the same Just take $b+c$ etc in the denominator That is write $\frac{(b+c)(a^4-b^2c^2)}{ab+2bc+ca}=\frac{a^4-b^2c^2}{a+\frac{2bc}{b+c}}$

About which solution are you saying?

for #1 \[\sum{\frac{(b+c)(a^4-b^2c^2)}{ab+2bc+ca}}=\] \[\sum{\frac{(a+b)(a^3b^3+3a^3b^2c+5a^3bc^2+3a^2b^3c+6a^2b^2c^2+5ab^3c^2+c^6)(a-b)^2}{(ab+ac+2bc)(ab+2ac+bc)(2ab+ac+bc)}}\]

apricot77 wrote: wlog \[a\geq b\geq c\] \[(b+c)(a^4-bc)-(a+c)(b^4-ac)\=\ab(a^3-b^3)+c(a^4-b^4)+c(a^2-b^2)+c^{2}(a-b)\] hence if \[a\geq b\geq c\] then \[(b+c)(a^4-bc)\geq (a+c)(b^4-ac)\geq (a+b)(c^4-ab)\] also \[ab+2bc+ca\leq ab+bc+2ca\leq 2ab+bc+ca\] By chebyshev inequality we obtain \[3\sum\frac{(b+c)(a^4-bc)}{ab+2bc+ca}\geq(\sum a^{4}b-\sum a^{3}b^{2})(\sum\frac{1}{ab+2bc+ca})\geq 0\] QED I was talking about this one