Let $a$, $b$ and $c$ be positive numbers. Prove that: $$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}+\frac{3\sqrt[3]{abc}}{2(a+b+c)}\leq2$$ I tried C-S, BW, uvw and more, but without any success.
Prove that $\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}+\frac{3\sqrt[3]{abc}}{2(a+b+c)}\leq2$
3
$\begingroup$
inequality
contest-math
-
0BW works here to solve the problem – 2017-01-18
-
0it is possible when you isolate the cube rrot! – 2017-01-18
-
0my mistake? or yours? – 2017-01-18
-
0all summands are positive in this case – 2017-01-18
-
0I proved this inequality!!! uvw helps here. Thanks all! – 2017-01-18
-
0post your solution please Michael – 2017-01-18
1 Answers
0
Here is my proof. Dedicated to dear Dr. Sonnhard Graubner.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$,$abc=w^3$ and $u=xw$.
Hence, $x\geq1$ and we need to prove that $$\sum_{cyc}\frac{a}{a+b}+\frac{w}{2u}\leq2$$ or $$\sum_{cyc}\left(\frac{a}{a+b}-\frac{1}{2}\right)\leq\frac{1}{2}-\frac{w}{2u}$$ or $$(u-w)(9uv^2-w^3)\geq u\sum_{cyc}(a-b)(a+c)(b+c)$$ or $$(u-w)(9uv^2-w^3)\geq u\sum_{cyc}(a-b)c^2$$ or $$(u-w)(9uv^2-w^3)\geq u(a-b)(a-c)(b-c),$$ which says that it remains to prove our inequality for $a\geq b\geq c$.
Id est, it remains to prove that $$(u-w)^2(9uv^2-w^3)^2\geq u^2(a-b)^2(a-c)^2(b-c)^2$$ or $$(u-w)^2(9uv^2-w^3)^2\geq 27u^2(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)$$ or $f(v^2)\geq0$, where $f(v^2)=108u^2v^6-81u^2w(2u-w)v^4-$
$-18uw^3(10u^2-2uw+w^2)v^2+w^3(108u^5+28u^2w^3-2uw^4+w^5)$.
But $$f'(v^2)=324u^2v^4-162u^2w(2u-w)v^2-18uw^3(10u^2-2uw+w^2),$$ which says that $v^2_{min}=\frac{6x^2-3x+\sqrt{36x^4+44x^3-7x^2+8x}}{12x}w^2$
and it remains to prove that $f\left(v^2_{min}\right)\geq0$ or $$648x^5-396x^4+342x^3+107x^2+20x+8\geq\sqrt{x(36x^3+44x^2-7x+8)^3}$$ or $$(x-1)^2(5832x^8+972x^7+2646x^6+1477x^5+581x^4+105x^3+51x^2-x+1)\geq0.$$ Done!