Prove that $\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}\geq1+4\sqrt{\frac{abcd}{(a+b)(b+c)(c+d)(d+a)}}$

Question

Let $a$, $b$, $c$ and $d$ be positive numbers. Prove that: $$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}\geq1+4\sqrt{\frac{abcd}{(a+b)(b+c)(c+d)(d+a)}}$$ We can prove this inequality by using $b=a+u$, $c=a+v$ and $d=a+w$ after squaring of the both sides. I am looking for a nice proof. Thank you!

Answer 1

Here is an easy proof. Let $x=\frac{b}{a}$, $y=\frac{c}{b}$, $z=\frac{d}{c}$ and $t=\frac{a}{d}$.

Hence, $xyzt=1$ and we need to prove that $\sum\limits_{cyc}\frac{1}{x+1}\geq1+\frac{4}{\sqrt{\prod\limits_{cyc}(x+1)}}$.

Let $f(x,y,z,t,\lambda)=\sum\limits_{cyc}\frac{1}{x+1}-1-\frac{4}{\sqrt{\prod\limits_{cyc}(x+1)}}+\lambda(xyzt-1)$.

It's obvious that in the minimum point of $f$ happens $\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=\frac{\partial f}{\partial z}=\frac{\partial f}{\partial t}=\frac{\partial f}{\partial \lambda}=0$.

But $\frac{\partial f}{\partial x}=-\frac{1}{(x+1)^2}+\frac{2}{(x+1)\sqrt{\prod\limits_{cyc}(x+1)}}+\frac{\lambda}{x}$, which gives $(x+1)\frac{\partial f}{\partial x}=-\frac{1}{x+1}+\frac{2}{\sqrt{\prod\limits_{cyc}(x+1)}}+\frac{\lambda}{x}+\lambda$. Thus, $-\frac{1}{x+1}+\frac{\lambda}{x}=-\frac{1}{y+1}+\frac{\lambda}{y}$, which gives $(x-y)\left(\frac{xy}{(x+1)(y+1)}-\lambda\right)=0$. Similarly we obtain:

$(x-z)\left(\frac{xz}{(x+1)(z+1)}-\lambda\right)=0$, $(x-t)\left(\frac{xt}{(x+1)(t+1)}-\lambda\right)=0$, $(y-z)\left(\frac{yz}{(y+1)(z+1)}-\lambda\right)=0$,

$(y-t)\left(\frac{yt}{(y+1)(t+1)}-\lambda\right)=0$ and $(z-t)\left(\frac{zt}{(z+1)(t+1)}-\lambda\right)=0$.

Since $\frac{xy}{(x+1)(y+1)}-\lambda=\frac{xz}{(x+1)(z+1)}-\lambda\Leftrightarrow y=z$,

we see that in the minimum point or two variables are equal twice or three variables are equal.

1. $y=x$ and $z=t=\frac{1}{x}$.

In this case $\sum\limits_{cyc}\frac{1}{x+1}\geq1+\frac{4}{\sqrt{\prod\limits_{cyc}(x+1)}}\Leftrightarrow(x+1)^2\geq4x$, which is obvious;

2. $y=z=x$ and $t=\frac{1}{x^3}$.

In this case $\sum\limits_{cyc}\frac{1}{x+1}\geq1+\frac{4}{\sqrt{\prod\limits_{cyc}(x+1)}}\Leftrightarrow(x-1)^2(9x^4+2x^3+5x^2+4x+4)$, which is obvious too.

Done!

Answer 2

Let: $ a+b+c+d=1 $

Then the inequality can be rewritten:

$$a\cdot \frac{c+d}{a+b}+b \cdot \frac{d+a}{b+c}+c \cdot \frac{a+b}{c+d}+d \cdot \frac{b+c}{d+a} \ge 4 \cdot \sqrt {\frac{abcd}{(a+b)(b+c)(c+d)(d+a)}}$$

$$\frac{a (c+d)^2+c (a+b)^2}{(a+b)(c+d)}+\frac{b (d+a)^2+d (b+c)^2}{(b+c)(d+a)} \ge 2 \cdot \sqrt {\frac{(a (c+d)^2+c (a+b)^2)(b (d+a)^2+d (b+c)^2)}{(a+b)(b+c)(c+d)(d+a)}} \ge$$ $$\ge 4\cdot \sqrt {\frac{abcd}{(a+b)(b+c)(c+d)(d+a)}}$$

$$\Leftrightarrow (a (c+d)^2+c (a+b)^2)(b (d+a)^2+d (b+c)^2) \ge 4abcd$$

$$4 abcd \sum\limits_{cyc} a^2+ \left ( 5 abcd\sum\limits_{cyc}ab+\sum\limits_{cyc}(a^4bc+ab^2d^3+ac^2d^3) \right) + $$

$$+\left ( 4abcd \cdot (ac+bd)+2 (a^3b^2c+ac^3d^2)+2 (b^3c^2d+bd^3a^2) \right) \ge 4abcd\cdot (a+b+c+d)^2= 4abcd$$