Prove that: $\sum\limits_{cyc}\frac{1}{a}\sum\limits_{cyc}\frac{1}{1+a^2}\geq\frac{16}{1+abcd}$

Question

Let $a$, $b$, $c$ and $d$ be positive numbers. Prove that: $$\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\left(\frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}+\frac{1}{1+d^2}\right)\geq\frac{16}{1+abcd}$$

I tried Rearrangement, C-S and more, but without success.

Answer 1

My third proof.

By Cauchy-Bunyakovsky-Schwarz inequality, we have $$\mathrm{LHS} \ge \left(\sum_{\mathrm{cyc}} \frac{1}{\sqrt{a(1+a^2)}}\right)^2.$$

It suffices to prove that $$\sum_{\mathrm{cyc}} \frac{1}{\sqrt{a(1+a^2)}} \ge \frac{4}{\sqrt{1 + abcd}}.$$

Using $2(1 + abcd) \ge (1 + \sqrt{abcd})^2$, it suffices to prove that $$\sum_{\mathrm{cyc}} \frac{1}{\sqrt{a(1+a^2)}} \ge \frac{4\sqrt 2}{1 + \sqrt{abcd}}$$ or $$\sum_{\mathrm{cyc}} \frac{1}{\sqrt{2a(1+a^2)}} \ge \frac{4}{1 + \sqrt{abcd}}. \tag{1}$$

Let $a = \mathrm{e}^x, b = \mathrm{e}^y, c = \mathrm{e}^z, d = \mathrm{e}^w$. We need to prove that $$\sum_{\mathrm{cyc}} \frac{1}{\sqrt{2\mathrm{e}^x(1+\mathrm{e}^{2x})}} \ge \frac{4}{1 + \mathrm{e}^{(x+y+z+w)/2}}. \tag{2}$$

Let $$f(x) := \frac{1}{\sqrt{2\mathrm{e}^x(1+\mathrm{e}^{2x})}}.$$ We have $$f''(x) = \frac{\sqrt 2\, (9\mathrm{e}^{4x} - 2\mathrm{e}^{2x} + 1)}{8(1+\mathrm{e}^{2x})^2\sqrt{\mathrm{e}^x(1+\mathrm{e}^{2x})}} > 0.$$ Thus, $f(x)$ is convex on $\mathbb{R}$.

By Jensen and AM-GM, we have $$\mathrm{LHS}_{(2)} \ge 4 f\left(\frac{x + y + z + w}{4}\right) = \frac{4}{\sqrt{2q(1+q^2)}} \ge \frac{4}{1 + q^2} = \mathrm{RHS}_{(2)}$$ where $q = \mathrm{e}^{(x+y+z+w)/4}$.

We are done.

Answer 2

My second solution

By Cauchy-Bunyakovsky-Schwarz inequality, we have \begin{align} &\frac{1}{1+a^2} + \frac{1}{1+b^2} + \frac{1}{1+c^2} + \frac{1}{1+d^2}\\ =\ & \frac{(1/a)^2}{1+(1/a)^2} + \frac{(1/b)^2}{1+(1/b)^2} + \frac{(1/c)^2}{1+(1/c)^2} + \frac{(1/d)^2}{1+(1/d)^2}\\ \ge\ & \frac{(1/a + 1/b + 1/c + 1/d)^2}{4 + (1/a)^2 + (1/b)^2 + (1/c)^2 + (1/d)^2}. \end{align} It suffices to prove that $$\frac{(1/a + 1/b + 1/c + 1/d)^3}{4 + (1/a)^2 + (1/b)^2 + (1/c)^2 + (1/d)^2}\ge \frac{16}{1 + abcd}$$ or equivalently $$\frac{(a+b+c+d)^3}{4 + a^2 + b^2 + c^2 + d^2} \ge \frac{16abcd}{1 + abcd}.\tag{1}$$

By Vasc's Equal Variable Theorem [1, Corollary 1.9], we only need to prove the case when $a=b=c \le d$. Let $d = a + s$ for $s \ge 0$. It suffices to prove that \begin{align} &a^3 s^4+(13 a^4-16 a^3+1) s^3+(60 a^5-48 a^4+12 a) s^2+(112 a^6-96 a^5-64 a^3+48 a^2) s\\ &\quad +64 a^7-64 a^6-64 a^4+64 a^3 \ge 0. \end{align} It is not difficult.

Remarks: I hope to see nice proofs for (1).

Reference

[1] Vasile Cirtoaje, “The Equal Variable Method”, J. Inequal. Pure and Appl. Math., 8(1), 2007. % https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf