Show that $a^2(a+b)^3+b^2(b+c)^3+c^2(c+a)^3\geq{\frac{8abc(a+b+c)^2}{3}}$ holds for $a,b,c>0$

Question

Let $a,b,c>0$. Show that $$a^2(a+b)^3+b^2(b+c)^3+c^2(c+a)^3\geq{\frac{8abc(a+b+c)^2}{3}}.$$

I have no idea. Please give a hint. Thanks!

Answer 1

The hint:

After full expanding we need to prove that $$\sum_{cyc}(3a^5+9a^4b+9a^3b^2+3a^3c^2-8a^3bc-16a^2b^2c)\geq0.$$ Now, prove by Rearrangement that $$\sum_{cyc}a^4b\geq\sum_{cyc}a^3bc$$ and the rest it's Rearrangement again or AM-GM.

Good luck!

Also, we can use Holder and C-S: $$\sum_{cyc}a^2(a+b)^3=\sum_{cyc}\frac{a^3(a+b)^3}{a}\geq\frac{\left(\sum\limits_{cyc}a(a+b)\right)^3}{3(a+b+c)}\geq$$ $$\geq\frac{\left(\frac{6}{9}\sum\limits_{cyc}(a^2+2ab)\right)^3}{3(a+b+c)}=\frac{8(a+b+c)^5}{81}\geq\frac{81abc(a+b+c)^2}{3}.$$

Answer 2

Proof

By expanding and rewriting, the inequality we want to prove is equivalent to

$3(a^5+b^5+c^5-a^2b^2c-a^2bc^2-ab^2c^2)+8(a^4b+b^4c+c^4a-a^3bc-ab^3c-abc^3)+(a^4b+b^4c+c^4a-a^2b^2c-a^2bc^2-ab^2c^2)+9(a^3b^2+b^3c^2+c^3a^2-a^2b^2c-a^2bc^2-ab^2c^2)+3(a^3c^2+b^3a^2+c^3b^2-a^2b^2c-a^2bc^2-ab^2c^2)\geq 0.$ $\tag0$

Step 1 Show that

$a^5+b^5+c^5 \geq a^2b^2c+a^2bc^2+ab^2c^2$ holds for $a,b,c>0$.

For this purpose, by Cauchy's Inequality, we have $$(bc+ca+ab)\left(\frac{a^4}{bc}+\frac{b^4}{ca}+\frac{c^4}{ab}\right)\geq (a^2+b^2+c^2)^2 \geq (ab+bc+ca)^2.\tag1$$ Thus,$$\frac{a^4}{bc}+\frac{b^4}{ca}+\frac{c^4}{ab} \geq ab+bc+ca.\tag2$$ As a result,$$a^5+b^5+c^5 \geq abc(ab+bc+ca)=a^2b^2c+a^2bc^2+ab^2c^2.\tag3$$

Step 2 Show that

$a^4b+b^4c+c^4a \geq a^3bc+ab^3c+abc^3$ holds for $a,b,c>0$.

For this purpose, by Cauchy's Inequality, we have $$(ca+ab+bc)\left(\frac{a^3}{c}+\frac{b^3}{a}+\frac{c^3}{b}\right)\geq (a^2+b^2+c^2)^2 \geq (ab+bc+ca)(a^2+b^2+c^2).\tag4$$ Thus,$$\frac{a^3}{c}+\frac{b^3}{a}+\frac{c^3}{b} \geq a^2+b^2+c^2.\tag5$$ As a result,$$a^4b+b^4c+c^4a \geq abc(a^2+b^2+c^2)=a^3bc+ab^3c+abc^3.\tag6$$

Step 3 Show that

$a^4b+b^4c+c^4a \geq a^2b^2c+a^2bc^2+ab^2c^2$ holds for $a,b,c >0$.

From $(6)$, we obtain$$a^4b+b^4c+c^4a \geq abc(a^2+b^2+c^2)\geq abc(ab+bc+ca)=a^2b^2c+a^2bc^2+ab^2c^2.\tag7$$

Step 4 Show that

$a^3b^2+b^3c^2+c^3a^2 \geq a^2b^2c+a^2bc^2+ab^2c^2$ holds for $a,b,c>0$.

For this purpose, by Cauchy's Inequality, we have $$(bc+ca+ab)\left(\frac{a^2b}{c}+\frac{b^2c}{a}+\frac{c^2a}{b}\right) \geq (ab+bc+ca)^2.\tag8$$ Thus,$$\frac{a^2b}{c}+\frac{b^2c}{a}+\frac{c^2a}{b} \geq ab+bc+ca.\tag9$$ As a result,$$a^3b^2+b^3c^2+c^3a^2\geq abc(ab+bc+ca)=a^2b^2c+a^2bc^2+ab^2c^2.\tag{10}$$

Step 5 Show that

$a^3c^2+b^3a^2+c^3b^2 \geq a^2b^2c+a^2bc^2+ab^2c^2$ holds for $a,b,c>0$.

For this purpose, by Cauchy's Inequality, we have $$(bc+ca+ab)\left(\frac{a^2c}{b}+\frac{b^2a}{c}+\frac{c^2b}{a}\right) \geq (ab+bc+ca)^2.\tag{11}$$ Thus,$$\frac{a^2c}{b}+\frac{b^2a}{c}+\frac{c^2b}{a} \geq ab+bc+ca.\tag{12}$$ As a result,$$a^3c^2+b^3a^2+c^3b^2\geq abc(ab+bc+ca)=a^2b^2c+a^2bc^2+ab^2c^2.\tag{13}$$

According to the five steps above, we can see that $(0)$ holds for $a,b,c>0.$

Please correct me if I'm wrong!

Answer 3

Another Proof

By Cauchy's Inequality, we have \begin{align*} &[(a+b)+(b+c)+(c+a)][a^2(a+b)^3+b^2(b+c)^3+c^2(c+a)^3]\\ \geq &[a(a+b)^2+b(b+c)^2+c(c+a)^2]^2,\end{align*}and \begin{align*} &(a+b+c)[a(a+b)^2+b(b+c)^2+c(c+a)^2]\\ \geq &[a(a+b)+b(b+c)+c(c+a)]^2.\end{align*}Besides, notice that $$a(a+b)+b(b+c)+c(c+a)\geq \frac{2}{3}(a+b+c)^2,$$ and $$(a+b+c)^3 \geq 27abc.$$ It follows that \begin{align*} a^2(a+b)^3+b^2(b+c)^3+c^2(c+a)^3 &\geq \frac{[a(a+b)^2+b(b+c)^2+c(c+a)^2]^2}{2(a+b+c)}\\ &\geq \frac{[a(a+b)+b(b+c)+c(c+a)]^4}{2(a+b+c)^3}\\ &\geq \frac{\left[\dfrac{2}{3}(a+b+c)^2\right]^4}{2(a+b+c)^3}\\&=\frac{8}{81}(a+b+c)^3(a+b+c)^2\\&\geq \frac{8}{81}\cdot 27abc \cdot (a+b+c)^2\\&=\frac{8}{3}abc(a+b+c)^2. \end{align*}