$ab(a+b) + bc(b+c) + ac(a+c) \geq \frac{2}{3}(a^{2}+b^{2}+c^{2})+ 4abc$ for $\frac1a+\frac1b+\frac1c=3$ and $a,b,c>0$

Question

Let $a$, $b$, and $c$ be positive real numbers with $\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3$. Prove that: $$ ab(a+b) + bc(b+c) + ac(a+c) \geq \frac{2}{3}(a^{2}+b^{2}+c^{2})+ 4abc. $$

Let us consider the following proofs. $$ a^{2}+b^{2}+c^{2} \geq ab+bc+ca $$

By the Arithmetic Mean-Geometric Mean Inequality we have $$ a^{2}+b^{2} \geq 2ab,\ \ b^{2}+c^{2} \geq 2bc,\ \ c^{2}+a^{2} \geq 2ca \tag{1} $$ If we add together all the inequalities $(1)$, we obtain $$ 2a^{2}+2b^{2}+2c^{2} \geq 2ab+2bc+2ca $$

By dividing both side by $2$, the result follows.

Now let us consider, $$ ab(a+b) + bc(b+c) + ac(a+c) \geq 6abc \tag{2} $$

I already have proved $(2)$

Then, We are given, $$ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3 \implies bc+ac+ab=3abc \tag{3} $$ Notice that we have $$ a^{2}+b^{2}+c^{2} \geq bc+ac+ab=3abc $$ So, $$ a^{2}+b^{2}+c^{2} \geq 3abc \tag{4} $$ Let us multiply both side of $(4)$ by $\displaystyle\frac{2}{3}$, yield $$ \frac{2}{3}(a^{2}+b^{2}+c^{2}) \geq 2abc $$

Here where I stopped. Would someone help me out ! Thank you so much

Answer 1

We note that \begin{align*} ab(a+b)+bc(b+c)+ac(a+c) &= a^2b+ab^2+b^2c+bc^2+a^2c+ac^2\\ &=\frac{a^2}{\frac{1}{b}}+\frac{b^2}{\frac{1}{a}} + \frac{b^2}{\frac{1}{c}}+ \frac{c^2}{\frac{1}{b}}+\frac{a^2}{\frac{1}{c}}+\frac{c^2}{\frac{1}{a}}\\ &\ge \frac{4(a+b+c)^2}{2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})} \qquad \mbox{(by the Schwarz inequality)}\\ &=\frac{2}{3}(a+b+c)^2\\ &=\frac{2}{3}(a^2+b^2+c^2+2ab+2bc+2ac)\\ &=\frac{2}{3}(a^2+b^2+c^2) + \frac{4}{3}(ab+bc+ca)\\ &=\frac{2}{3}(a^2+b^2+c^2) + 4abc. \end{align*} Here, we employed the Schwarz inequality of the from \begin{align*} &\ (a+b+b+c+a+c)^2 \\ =&\ \left(\frac{a}{\sqrt{\frac{1}{b}}}\sqrt{\frac{1}{b}}+\frac{b}{\sqrt{\frac{1}{a}}}\sqrt{\frac{1}{a}}+\frac{b}{\sqrt{\frac{1}{c}}}\sqrt{\frac{1}{c}}+\frac{c}{\sqrt{\frac{1}{b}}}\sqrt{\frac{1}{b}}+\frac{a}{\sqrt{\frac{1}{c}}}\sqrt{\frac{1}{c}}+\frac{c}{\sqrt{\frac{1}{a}}}\sqrt{\frac{1}{a}}\right)^2\\ \le&\ \left(\frac{a^2}{\frac{1}{b}}+\frac{b^2}{\frac{1}{a}} + \frac{b^2}{\frac{1}{c}}+ \frac{c^2}{\frac{1}{b}}+\frac{a^2}{\frac{1}{c}}+\frac{c^2}{\frac{1}{a}}\right)\left(2\big(\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \big)\right). \end{align*}

Answer 2

We need to prove that $$\sum\limits_{cyc}(a^2b+a^2c)\geq\frac{2(a^2+b^2+c^2)}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}+4abc$$ or $$\sum\limits_{cyc}(a^2b+a^2c)\geq\frac{2abc(a^2+b^2+c^2)}{ab+ac+bc}+4abc$$ or $$\sum\limits_{cyc}ab\sum\limits_{cyc}(a^2b+a^2c)\geq2abc(a^2+b^2+c^2)+4abc(ab+ac+bc)$$ or $$\sum_{cyc}(a^3b^2+a^3c^2+2a^3bc+2a^2b^2c)\geq\sum\limits_{cyc} (2a^3bc+4a^2b^2c)$$ $$\sum\limits_{cyc}(a^3b^2+a^3c^2-2a^2b^2c)\geq0$$ or $$\sum\limits_{cyc}c^2(a+b)(a-b)^2\geq0$$ Done!