Find minimum value of $7(a^4+b^4+c^4)+\frac{ab+bc+ca}{a^2b+b^2c+c^2a}$

Question

Let $a,b,c\in \mathbb{R}^+$ such that $a+b+c=3$. Find minimum value of $$7(a^4+b^4+c^4)+\dfrac{ab+bc+ca}{a^2b+b^2c+c^2a}$$

I tried to use $(a^2+b^2+c^2)(a+b+c)\geq3(a^2b+b^2c+c^2a)$, but without success.

Answer 1

I think your idea works! $$(a^2+b^2+c^2)(a+b+c)\geq3(a^2b+b^2c+c^2a)$$ it's $$\sum_{cyc}(a^3-2a^2b+a^2c)\geq0$$ or $$\sum_{cyc}(a^3-2a^2b+ab^2)\geq0$$ or $$\sum_{cyc}a(a-b)^2\geq0,$$ which is obvious.

Thus, $$a^2b+b^2c+c^2a\leq a^2+b^2+c^2.$$

Now, for $a=b=c=1$ we get a value $22$.

We'll prove that it's a minimal value.

Indeed, by your lemma it's enough to prove that $$7(a^4+b^4+c^4)+\frac{ab+ac+bc}{a^2+b^2+c^2}\geq22.$$ By AM-GM $$\sum_{cyc}(a^4+3)\geq\sum_{cyc}4\sqrt[4]{a^4\cdot1\cdot1\cdot1}=4(a+b+c)=12.$$ Thus, $$a^4+b^4+c^4\geq3.$$ Also, by AM-GM again we obtain: $$\sum_{cyc}(a^4+1)\geq\sum_{cyc}2\sqrt{a^4\cdot1}=2(a^2+b^2+c^2).$$ Id est, it's enough to prove that $$a^2+b^2+c^2+\frac{ab+ac+bc}{a^2+b^2+c^2}\geq4$$ or $$\frac{9(a^2+b^2+c^2)}{(a+b+c)^2}+\frac{ab+ac+bc}{a^2+b^2+c^2}\geq4.$$ Let $a^2+b^2+c^2=k(ab+ac+bc).$

Thus, $k\geq1$ and we need to prove that $$\frac{9k}{k+2}+\frac{1}{k}\geq4$$ or $$(k-1)(5k-2)\geq0,$$ which is obvious.

Done!