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

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

• Yes, I am sure.Thank you ! – ADAM Sep 27 '16 at 20:11

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*}

• Which mean have you used to come up with that $$\frac{a^2}{\frac{1}{b}}+\frac{b^2}{\frac{1}{a}} + \frac{b^2}{\frac{1}{c}}+ \frac{c^2}{\frac{1}{b}}+\frac{a^2}{\frac{1}{c}}+\frac{c^2}{\frac{1}{a}} \ge \frac{4(a+b+c)^2}{2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}$$ – ADAM Sep 30 '16 at 19:54
• @ADAM: see updates above. – Gordon Sep 30 '16 at 20:09
• I would assume that inequality is right. $$\left(\frac{a^2}{\frac{1}{b}}+\frac{b^2}{\frac{1}{a}} + \frac{b^2}{\frac{1}{c}}+ \frac{c^2}{\frac{1}{b}}+\frac{a^2}{\frac{1}{c}}+\frac{c^2}{\frac{1}{a}}\right)\left(2\big(\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \big)\right) \geq 4(a^2+b^2+c^2)+8(ab+ac+bc)$$ so, you are done? they only thing left is $$ab(a+b)+bc(b+c)+ac(a+c) = a^2b+ab^2+b^2c+bc^2+a^2c+ac^2 =\frac{a^2}{\frac{1}{b}}+\frac{b^2}{\frac{1}{a}} + \frac{b^2}{\frac{1}{c}}+ \frac{c^2}{\frac{1}{b}}+\frac{a^2}{\frac{1}{c}}+\frac{c^2}{\frac{1}{a}}$$ Is that right? – ADAM Sep 30 '16 at 21:48
• Would you please past the link of Schwarz inequality form that you have used in your proof ! Thank you ! – ADAM Sep 30 '16 at 23:56
• The general Schwarz inequality has the form $|\sum_{i=1}^nx_iy_i |\le (\sum_{i=1}^n x_i^2)^{1/2}(\sum_{i=1}^n y_i^2)^{1/2}$. I think it is already there. – Gordon Oct 1 '16 at 0:53

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!

• If I conider the first one , which is $$(a^2b+a^2c)+(b^2c+b^2a)+(c^2a+c^2b)$$. Would you explain how it this help me out ? – ADAM Sep 28 '16 at 17:48
• would you please give me any other hint! Thank you – ADAM Sep 28 '16 at 20:35
• @ADAM I don't understand your question. It's just full expanding. – Michael Rozenberg Sep 28 '16 at 21:03
• if I expand your work on the LHS it will come back as the original problem expat is different ordering. On the RHS you have replaced the 3 with $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3$$ – ADAM Sep 28 '16 at 21:12
• $$\sum\limits_{cyc}(a^2b+a^2c)=(a^2b+a^2c)+(b^2c+b^2a)+(c^2a+c^2b)$$ – ADAM Sep 28 '16 at 21:45