Prove using Jensen's inequality that if $abcd=1$ then $\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}+\frac{1}{(1+c)^{2}}+\frac{1}{(1+d)^{2}} \geq 1$

Question

Question -

Let $a, b, c, d$ be positive real numbers such that abcd $=1 .$ Prove that $$ \begin{array}{c} \frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}+\frac{1}{(1+c)^{2}}+\frac{1}{(1+d)^{2}} \geq 1 \\ \text { (China TST 2004) } \end{array} $$

I solved it using CS inequality, and also seen a solution using the fact that $\frac{1}{(1+x)^{2}}+\frac{1}{(1+y)^{2}} \geq \frac{1}{1+x y}$

but, i think the inequality is very much likely to be also solved using jensens but i am not able to find proof, i tried many functions such as $x^2 , 1/x^2 , 1/(x+1)^2$ but none of them working...

can anyone solve it using Jensen inequality ???

Answer 1

We can use the Vasc's RCF Theorem. It's like Jensen, but it's not Jensen.

Also, since $f(x)=\frac{1}{(1+e^x)^2}$ has an unique inflection point, we can use Jensen with Karamata, but it's not so nice solution. I am ready to show, if you want.

Indeed, $$f''(x)=\frac{4e^x\left(e^x-\frac{1}{2}\right)}{(1+e^x)^4}.$$ Thus, $f$ is a convex function on $[-\ln2,+\infty)$ and a concave function on $(-\infty,-\ln2]$.

We need to prove that $$\sum_{cyc}f(x)\geq0,$$ where $x+y+z+t=0$.

Now, let $x\geq y\geq z\geq t.$

We'll consider the following cases.

1. $x\geq y\geq z\geq t\geq-\ln2.$

Thus, by Jensen $$\sum_{cyc}f(x)\geq4f\left(\frac{x+y+z+t}{4}\right)=4f(0)=1.$$

2. $x\geq y\geq z\geq-\ln2\geq t$.

Thus, by Jensen again: $$\sum_{cyc}f(x)\geq3f\left(\frac{x+y+z}{3}\right)+f(t)=3f\left(\frac{-t}{3}\right)+f(t).$$ Thus, it's enough to prove that $$3f\left(\frac{-t}{3}\right)+f(t)\geq0,$$ which is $$\sum_{cyc}\frac{1}{(1+a)^2}\geq1,$$ where $b=c=a$ and $d=\frac{1}{a^3}$ or $$\frac{3}{(1+a)^2}+\frac{1}{\left(1+\frac{1}{a^3}\right)^2}\geq1$$ or $$(a-1)^2(3a^2-2a+2)\geq0,$$ which is obvious.

3. $x\geq y\geq-\ln2\geq z\geq t$.

Thus, by Jensen again we have: $$f(x)+f(y)\geq2f\left(\frac{x+y}{2}\right)^2=\frac{2}{\left(1+e^{\frac{x+y}{2}}\right)^2}=\frac{2}{(1+\sqrt{ab})^2}.$$

Also, since $$(-\ln2,\ln2+z+t)\succ(z,t),$$ by Karamata we obtain: $$f(z)+f(t)\geq f(-\ln2)+f(\ln2+z+t)=$$ $$=\frac{1}{\left(1+e^{-\ln2}\right)^2}+\frac{1}{\left(1+e^{\ln+z+t}\right)^2}=\frac{4}{9}+\frac{1}{(1+2cd)^2}.$$ Let $\sqrt{ab}=u$.

Thus, it's enough to prove in this case that $$\frac{4}{9}+\frac{1}{\left(1+\frac{2}{u^2}\right)^2}+\frac{2}{(1+u)^2}\geq1$$ and since $$\left(\frac{1}{\left(1+\frac{2}{u^2}\right)^2}+\frac{2}{(1+u)^2}\right)'=\frac{4(u^3-2)(u^3+6u^2+4)}{(u+1)^3(u^2+2)^3},$$ it's enough to prove the last inequality for $u=\sqrt[3]2,$ which gives $$\frac{4}{9}+\frac{1}{\left(1+\frac{2}{u^2}\right)^2}+\frac{2}{(1+u)^2}=\frac{4}{9}+\frac{1}{(1+\sqrt[3]2)^2}+\frac{2}{(1+\sqrt[3]2)^2}>\frac{4}{9}+\frac{3}{(1+1.3)^2}>1.$$

4. $x\geq-\ln2\geq y\geq z\geq t$.

Thus, since $$\left(-\ln2,-\ln2,2\ln2+y+z+t\right)\succ(y,z,t),$$ by Karamata again we obtain: $$f(y)+f(z)+f(t)\geq2f(-\ln2)+f(2\ln2+y+z+t)=$$ $$=\frac{8}{9}+\frac{1}{\left(1+e^{2\ln2+y+z+t}\right)^2}=\frac{8}{9}+\frac{1}{(1+4bcd)^2}.$$ Id est, it's enough to prove that: $$\frac{1}{(1+a)^2}+\frac{8}{9}+\frac{1}{\left(1+\frac{4}{a}\right)^2}\geq1$$ or $$8a^4+8a^3-15a^2+32a+128\geq0,$$ which is obvious.

The case $-\ln2\geq x\geq y\geq z\geq t$ is impossible and we are done!