# Prove $\sum\limits_{cyc}\sqrt[3]{\frac{a^2+bc}{b+c}}\ge\sqrt[3]{9(a+b+c)}$

Let $a,b,c$ be positive real numbers. Show that $$\sum_{cyc}\sqrt[3]{\dfrac{a^2+bc}{b+c}}\ge\sqrt[3]{9(a+b+c)}$$

I have tried C-S and Holder inequalities, without success. How to solve it?

• This question is missing context or other details: Please improve the question by providing additional context, which ideally includes the source of the problem, its motivation, your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. Posts that merely state a question for solving are discouraged on this site. – Carl Mummert Feb 27 '18 at 19:53

The hint.

By Holder $$\left(\sum\limits_{cyc}\sqrt[3]{\frac{a^2+bc}{b+c}}\right)^3\sum_{cyc}(b+c)(a^2+bc)^3(3a+4b+4c)^4\geq\left(\sum_{cyc}(a^2+bc)(3a+4b+4c)\right)^4.$$ Thus, it remains to prove that $$\left(\sum_{cyc}(a^2+bc)(3a+4b+4c)\right)^4\geq9(a+b+c)\sum_{cyc}(b+c)(a^2+bc)^3(3a+4b+4c)^4,$$ which is true by $uvw$ (this inequality is equivalent to $f(w^3)\geq0,$ where $f$ is a cocave function).

About $uvw$ see here: https://math.stackexchange.com/tags/uvw/info

https://artofproblemsolving.com/community/c6h278791

Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2,$ where $v>0$, and $abc=w^3$.

Thus, we need to prove that $f(w^3)\geq0$, where $$f(w^3)=(27u^3-3uv^2-2w^3)^4+u(A(u,v^2)w^9+B(u,v^2)w^6+C(u,v^2)w^3+D(u,v^2)),$$ where $$A(u,v^2)=6042u^2-52v^2,$$ $$B(u,v^2)=-232389u^5+104382u^3v^2-5049uv^4,$$ $$C(u,v^2)=177147u^8+8748u^6v^2+78489u^4v^4-64881u^2v^6+459v^8$$ and $$D(u,v^2)=-1594323u^9v^2+1948617u^7v^4-618921u^5v^6-48114u^3v^8+5589uv^{10}.$$ Now, since $u\geq v\geq w$, we obtain: $$f''(w^3)=48(27u^3-3uv^2-2w^3)^2+6uA(uv^2)w^3+2uB(u,v^2)=$$ $$=48(27u^3-3uv^2-2w^3)^2+6u(6042u^2-52v^2)w^3+$$ $$+2u(-232389u^5+104382u^3v^2-5049uv^4)=$$ $$=-429786u^5+200988u^4v^2-9666u^2v^4+31068u^3w^3+264uv^2w^3+192w^6\leq0,$$ which says that $f$ is a concave function.

But a concave function gets a minimal value for an extreme value of $w^3$,

which says that it's enough to prove our inequality for an extreme value of $w^3$.

Now, $a$, $b$ and $c$ they are positive roots of the following equation. $$(x-a)(x-b)(x-c)=0$$ or $$x^3-(a+b+c)x^2+(ab+ac+bc)x-abc=0$$ or $$w^3=x^3-3ux^2+3v^2x,$$ which says that the line $y=w^3$ and a graph of $f$,

where $f(x)=x^3-3ux^2+3v^2x$, have three common points (with maybe a degree of the point).

Now, $$f'(x)=3(x^2-2ux+v^2),$$ which says that $x_{max}=u-\sqrt{u^2-v^2}$ and $x_{min}=u+\sqrt{u^2-v^2}$ and we see that

$w^3$ gets a maximal value, when the line $y=w^3$ is a tangent line to the graph of $f$,

which happens for equality case of two variables.

Also, we see that $f$ gets a minimal value, when the line $y=w^3$ is a tangent line to the graph of $f$,

which happens for equality case of two variables or we need to check what happens for $w^3\rightarrow0^+$.

Id est, it's enough to prove that $f(w^3)\geq0$ in the following cases:

1. $b=c$.

Since this inequality is homogeneous we can assume $b=c=1$, which gives: $$((a^2+1)(3a+8)+2(a+1)(4a+7))^4\geq$$ $$\geq9(a+2)(2(a^2+1)^3(3a+8)^4+2(a+1)^4(4a+7)^4)$$ or $$(a-1)^2(81a^{10}+432a^9-1161a^8-10320a^7-2093a^6+115994a^5+361653a^4+525718a^3+437016a^2+184086a+364)\geq0,$$ which is true for all $a>0$;

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

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

Thus, we need to prove that: $$(a^2(3a+4)+(4a+3)+a(4a+4))^4\geq$$ $$\geq9(a+1)(a^6(3a+4)^4+a(4a+3)^4+(a+1)a^3(4a+4)^4)$$ or $$(a+1)^2(81a^{10}-27a^9-324a^8+87a^7+214a^6+29a^5+214a^4+87a^3-324a^2-27a+81)\geq0,$$ which is true.

• can you post detail solution?or have other methods? – function sug Feb 27 '18 at 23:48
• Can someone explain me why he down voted? – Michael Rozenberg Feb 28 '18 at 3:33
• @function sug I added something. See now. – Michael Rozenberg Mar 4 '18 at 13:53

Just a little hint

By AM GM $$\sum \sqrt[3] {\frac {a^2+bc}{b+c}}\ge 3\sqrt[3] {\prod \left (\frac {a^2}{b+c}+ \frac {bc}{b+c}\right)^{\frac {1}{3}}}$$

Now by Hölder's inequality $$3\sqrt[3] {\prod \left(\frac {a^2}{b+c}+ \frac {bc}{b+c}\right)^{\frac {1}{3}}}\ge 3\sqrt[3] {\left (\frac {a^2b^2c^2}{(a+b)(b+c)(a+c)}\right)^{\frac {1}{3}}+\left (\frac {a^2b^2c^2}{(a+b)(b+c)(a+c)}\right)^{\frac {1}{3}}}=3\sqrt[3] {2\sqrt[3] {\frac {a^2b^2c^2}{(a+b)(b+c)(a+c)}}}$$

Now again by AM GM on denominator

$$3\sqrt[3] {2\sqrt[3] {\frac {a^2b^2c^2}{(a+b)(b+c)(a+c)}}}\ge 3\sqrt[3] {\frac {3}{(a+b+c)}\sqrt[3] {a^2b^2c^2}}$$

Hope it helped any way