Help to prove $\frac{a^3b+b^3c+c^3a}{a+b+c}+1\ge 2\sqrt{\frac{a^2b^2+b^2c^2+c^2a^2}{a+b+c}}$.

Question

I deal with a problem belonged RMM magazine without success. It's

Let $a,b,c>0: abc=1$. Prove that: $$\frac{a^3b+b^3c+c^3a}{a+b+c}+1\ge 2\sqrt{\frac{a^2b^2+b^2c^2+c^2a^2}{a+b+c}}$$

My idea is proving $a^3b+b^3c+c^3a\ge a^2b^2+b^2c^2+c^2a^2.$

If my statement is true, we'll use AM-GM

$$\frac{a^3b+b^3c+c^3a}{a+b+c}+1\ge \frac{a^2b^2+b^2c^2+c^2a^2}{a+b+c}+1\ge 2\sqrt{\frac{a^2b^2+b^2c^2+c^2a^2}{a+b+c}}$$ Thus, $$a^3b+b^3c+c^3a\ge a^2b^2+b^2c^2+c^2a^2 \iff \frac{a^2}{c}+\frac{c^2}{b}+\frac{b^2}{a}\ge \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}.$$ By C-S, $$\frac{a^2}{c}+\frac{c^2}{b}+\frac{b^2}{a}\ge \frac{(a+b+c)^2}{a+b+c}=a+b+c$$But $$a+b+c\ge \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b} \iff abc(a+b+c)\ge a^2b^2+b^2c^2+c^2a^2 $$ $$(ab-bc)^2+(bc-ca)^2+(ca-ab)^2\le 0.$$Can you help me prove this inequality? I'm appreciate your help.

Answer 1

Proof.

By multiplying $a+b+c$ and use $abc=1,$ we'll prove $$\frac{a^2}{c}+\frac{c^2}{b}+\frac{b^2}{a}+a+b+c\ge 2\sqrt{\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\right)(a+b+c)}$$or$$\frac{b}{ac}(bc+ac)+\frac{c}{ab}(ab+ca)+\frac{a}{bc}(bc+ab)\ge 2\sqrt{\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\right)(a+b+c)}.$$Now, we may use Cauchy-Schwarz inequality as $$\frac{b}{ac}(bc+ac)+\frac{c}{ab}(ab+ca)+\frac{a}{bc}(bc+ab)$$ $$=\left(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\right)\left(ab+bc+ca\right)-\frac{b^2}{c}-\frac{c^2}{a}-\frac{a^2}{b}$$ $$=\sqrt{\left[\frac{a^2}{b^2c^2}+\frac{b^2}{c^2a^2}+\frac{c^2}{a^2b^2}+2\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\right]\left(a^2c^2+b^2a^2+b^2c^2+2abc(a+b+c)\right)}-\frac{b^2}{c}-\frac{c^2}{a}-\frac{a^2}{b}$$ $$\ge 2\sqrt{abc(a+b+c)\cdot\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)}=2\sqrt{\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\right)(a+b+c)}.$$ Hence, the proof is done. Equality holds at $a=b=c=1.$

Answer 2

We need to prove that: $$a^3b+b^3c+c^3a+abc(a+b+c)\geq2\sqrt{(a+b+c)(a^2b^2+a^2c^2+b^2c^2)abc}$$ or $$\sum_{cyc}(a^3b+a^3c+2a^2bc)-4\sqrt{(a+b+c)(a^2b^2+a^2c^2+b^2c^2)abc}\geq\sum_{cyc}(a^3c-a^3b)$$ or $$\sum_{cyc}(a^3b+a^3c+2a^2bc)-4\sqrt{(a+b+c)(a^2b^2+a^2c^2+b^2c^2)abc}\geq(a+b+c)(a-b)(b-c)(c-a)$$ and since by AM-GM twice $$\sum_{cyc}(a^3b+a^3c+2a^2bc)\geq\sum_{cyc}(2a^2b^2+2a^2bc)\geq4\sqrt{(a+b+c)(a^2b^2+a^2c^2+b^2c^2)abc},$$ we can assume that $a\geq c\geq b.$

Now, let $c=b+u$ and $a=b+u+v$, $u=xv$, where $u$, $v$ and $x$ are non-negatives.

Thus, we need to prove that: $$\left(\sum_{cyc}(a^3b+a^2bc)\right)^2\geq4abc(a+b+c)(a^2b^2+a^2c^2+b^2c^2)$$ or $$12(u^2+uv+v^2)b^6+4(11u^3+12u^2v+15uv^2+7v^3)b^5+$$ $$+4(16u^4+17u^3v+23u^2v^2+22uv^3+6v^4)b^4+$$ $$+8(6u^5+5u^4v+5u^3v^2+10u^2v^3+6uv^4+v^5)b^3+$$ $$+(21u^6+10u^5v-15u^4v^2+10u^3v^3+22u^2v^4+8uv^5+v^6)b^2+$$ $$+2u^3(3u^4-5u^2v^2-5uv^3-v^4)b+u^6(u+v)^2\geq0,$$ for which it's enough to prove that: $$(21x^6+10x^5-15x^4+10x^3+22x^2+8x+1)(x+1)^2\geq(3x^4-5x^2-5x-1)^2$$ or $$x^2(3x^6+13x^5+14x^4+5x^3+2x^2+3x+1)\geq0$$ and we are done!