Explanation to an unmotivated step in an inequality.

Question

Follows the original problem:

Let $a, b, c$ be non-negative real numbers. Prove that $$ \frac{a^2}{a^2 + 2\left(a + b\right)^2} + \frac{b^2}{b^2 + 2\left(b + c\right)^2} + \frac{c^2}{c^2 + 2\left(c + a\right)^2} \geqslant \frac{1}{3} $$

This is my approach:
The inequality is equivalent to $$ \frac{1}{1 + 2\left(1 + \frac{b}{a}\right)^2} + \frac{1}{1 + 2\left(1 + \frac{c}{b}\right)^2} + \frac{1}{1 + 2\left(1 + \frac{a}{c}\right)^2} \geqslant \frac{1}{3} $$ Hence, Let $x$ be $\displaystyle \frac{b}{a}$, $y$ be $\displaystyle \frac{c}{b}$ and $z$ be $\displaystyle \frac{a}{c}$. Now the inequality is $$ \frac{1}{1 + 2\left(1 + x\right)^2} + \frac{1}{1 + 2\left(1 + y\right)^2} + \frac{1}{1 + 2\left(1 + z\right)^2} \geqslant \frac{1}{3} \quad\textrm{with}\quad xyz = 1 $$ $$ \frac{xyz}{xyz + 2(x + xyz)^2} + \frac{xyz}{xyz + 2(y + xyz)^2} + \frac{xyz}{xyz + 2(z + xyz)^2} \geqslant \frac{1}{3} $$ $$ \frac{xyz}{xyz + 2x^2(1 + yz)^2} + \frac{xyz}{xyz + 2y^2(1 + zx)^2} + \frac{xyz}{xyz + 2z^2(1 + xy)^2} \geqslant \frac{1}{3} $$ $$ \frac{yz}{yz + 2x(1 + yz)^2} + \frac{zx}{zx + 2y(1 + zx)^2} + \frac{xy}{xy + 2z(1 + xy)^2} \geqslant \frac{1}{3} $$ $$ \frac{y^2z^2}{y^2z^2 + 2(1 + yz)^2} + \frac{z^2x^2}{z^2x^2 + 2(1 + zx)^2} + \frac{x^2y^2}{x^2y^2 + 2(1 + xy)^2} \geqslant \frac{1}{3} $$ By $T_2$'s Lemma, $$ \frac {\left(xy + yz + zx\right)^2} {x^2y^2 + y^2z^2 + z^2x^2 + 2\left[(1 + yz)^2 + (1 + zx)^2 + (1 + xy)^2\right]} \geqslant \frac{1}{3} $$ Hence, after simplification, $$ 3(x + y + z) \geqslant 2(xy + yz + zx) + 3 $$ Substitute back $a,b,c$. $$ 3\left( \frac {a} {b} + \frac {b} {c} + \frac {c} {a} \right) \geqslant 2\left( \frac{a}{c} + \frac{b}{a} + \frac{c}{b} \right) + 3 $$ I think that it looks easy, but if someone has a way, it might become a solution.

Book's Approach:
It is the same till the second inequality in my approach, but then without any motivation goes on to assume $$x = \frac{np}{m^2}, y = \frac{mp}{n^2}, z = \frac{mn}{p^2}.$$ And then it is just $T_2$'s Lemma and some simplification to get squares on the greater side.
I don't see anything natural in assuming that and neither there is a motivation to take such step, besides it gives proof.
But my question is how does the author assume such a thing? And is there a way to complete my proof?

Thanks!

Answer 1

The $$3\left( \frac {a} {b} + \frac {b} {c} + \frac {c} {a} \right) \geqslant 2\left( \frac{a}{c} + \frac{b}{a} + \frac{c}{b} \right) + 3, \quad (*)$$ does not hold for $a = \frac 12$, $b = \frac 16,$ $c = 1.$

We need to prove $$\frac{1}{1 + 2\left(1 + x\right)^2} + \frac{1}{1 + 2\left(1 + y\right)^2} + \frac{1}{1 + 2\left(1 + z\right)^2} \geqslant \frac{1}{3}, \quad xyz=1.$$ Let $(x,\,y,\,z) =\left(\dfrac{bc}{a^2},\,\dfrac{ca}{b^2},\,\dfrac{ab}{c^2}\right)$ the inequality become $$\sum \frac{a^4}{3a^4+2b^2c^2+4a^2bc} \geqslant \dfrac{1}{3}.$$ Using the Cauchy-Schwarz inequality, we have $$\sum \frac{a^4}{3a^4+2b^2c^2+4a^2bc} \geqslant \dfrac{(a^2+b^2+c^2)^2}{3(a^4+b^4+c^4)+2(a^2b^2+b^2c^2+a^2c^2)+4abc(a+b+c)}$$ $$\geqslant \frac{(a^2+b^2+c^2)^2}{3(a^4+b^4+c^4)+2(a^2b^2+b^2c^2+c^2a^2)+4(a^2b^2+b^2c^2+c^2a^2)}=\frac{1}{3}.$$

Note. The inequality $(*)$ is true if $a,\,b,\,c$ the length of the sides of the triangle.

Answer 2

Your last inequality is wrong.

Try $b=4$, $c=2$ and $a\rightarrow+\infty$.

By the way, we can prove your inbequality by C-S, AM-GM and SOS: $$\sum_{cyc}\frac{a^2}{a^2+2(a+b)^2}=\sum_{cyc}\frac{a^2(a+c)^2}{(3a^2+4ab+2b^2)(a+c)^2}\geq$$ $$\geq\frac{\left(\sum\limits_{cyc}(a^2+ab)\right)^2}{\sum\limits_{cyc}(3a^2+4ab+2b^2)(a+c)^2}\geq\frac{1}{3},$$ where the last inequality it's $$\sum_{cyc}(a^3b+a^2b^2-2a^2bc)\geq0,$$ which is true by AM-GM and SOS: $$\sum_{cyc}(a^3b+a^2b^2-2a^2bc)=\sum_{cyc}(a^3b+c^2ab+a^2b^2-3a^2bc)\geq$$ $$\geq\sum_{cyc}\left(2\sqrt{a^3b\cdot c^2ab}+a^2b^2-3a^2bc\right)=\sum_{cyc}(a^2b^2-a^2bc)=\frac{1}{2}\sum_{cyc}c^2(a-b)^2\geq0.$$