# proving :$\frac{(ab+b)(2b+1)}{(ab+a)(5b+1)}+\frac{(bc+c)(2c+1)}{(bc+b)(5c+1)}+\frac{(ca+a)(2a+1)}{(ca+c)(5a+1)}\ge\frac{3}{2}$

Let $a,b,c, >0$ be real numbers such that $$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge3$$

How to prove that :

$$\frac{(ab+b)(2b+1)}{(ab+a)(5b+1)}+\frac{(bc+c)(2c+1)}{(bc+b)(5c+1)}+\frac{(ca+a)(2a+1)}{(ca+c)(5a+1)}\ge\frac{3}{2}$$

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I think there is some problem with first and second term. Match them with the pattern of third term. –  Aang Jul 12 '12 at 17:11
I don't see any problem. –  Host-website-on-iPage Jul 12 '12 at 17:14
The title didn't match the statement in the body. I'll edit. –  Robert Israel Jul 12 '12 at 17:14
now it's fine :) –  Aang Jul 12 '12 at 17:19
It comes from here fen.bilkent.edu.tr/~cvmath/Problem/problem_2011.htm (April 2011) –  Unoqualunque Sep 2 '12 at 11:06

I) Let us define for $x\geq 0$ the rational function

$$\tag{1} f(x)~:=~ \frac{(x+1)(x+5)}{x+2}~=~ x+4 - \frac{3}{x+2}~>~0, \qquad x\geq 0.$$

It is not hard to see that $f$ is positive, monotonically increasing

$$\tag{2} f'(x)~=~ 1 + \frac{3}{(x+2)^2}~>~0, \qquad x\geq 0,$$

and concave for $x\geq 0$. The tangent at $x=1$ is above the concave function,

$$\tag{3} f(x) ~\leq~ f(1) +(x-1) f'(1)~=~\frac{4}{3}(x+2), \qquad x\geq 0.$$

II) By going to reciprocal variables OP's inequality (v7) can be written as

$$\tag{4} \forall x,y,z\geq 0:\quad x^2 +y^2 +z^2 \geq 3 \quad\Rightarrow\quad \frac{x+1}{f(y)} + \frac{y+1}{f(z)} + \frac{z+1}{f(x)} ~\stackrel{?}{\geq}~ \frac{3}{2}.$$

III) If we could prove that

$$\tag{5} \forall x,y,z\geq 0:\quad x^2 +y^2 +z^2 \geq 3 \quad\Rightarrow\quad \frac{x+1}{y+2} + \frac{y+1}{z+2} + \frac{z+1}{x+2} ~\stackrel{?}{\geq}~ 2,$$

then eq. (4) would follow from eqs. (3) and (5).

IV) Equation (5) can be rewritten as

$$\tag{6} \forall x,y,z\geq 0:\quad x^2 +y^2 +z^2 \geq 3 \quad\Rightarrow \quad g(x,y,z)~\stackrel{?}{\geq}~ 0,$$

where

$$g(x,y,z) ~:=~-2(x+2)(y+2)(z+2)+ \sum_{{\rm cycl.}~ x,y,z}(x+1)(x+2)(y+2)$$ $$~=~ -2xyz -4 +\sum_{{\rm cycl.}~ x,y,z}\left(x^2y+2x^2-xy\right)$$ $$\tag{7} ~=~ 2\underbrace{\left(-xyz+\frac{1}{3}\sum_{{\rm cycl.}~ x,y,z}x^2y\right)}_{\geq 0 ~\text{because of (AG)}.} +\underbrace{\left(-3+\sum_{{\rm cycl.}~ x,y,z}x^2\right)}_{\geq 0}+h(x,y,z),$$

where

$$\tag{8} h(x,y,z) ~:=~ -1+\sum_{{\rm cycl.}~ x,y,z}\left(\frac{1}{3}x^2y+x^2-xy\right).$$

In equation (7) we have used the inequality of arithmetic and geometric means (AG).

V) If we could prove that

$$\tag{9} \forall x,y,z\geq 0:\quad x^2 +y^2 +z^2 \geq 3 \quad\Rightarrow \quad h(x,y,z)~\stackrel{?}{\geq}~ 0,$$

then eq. (6) would follow from eqs. (7) and (9).

VI) The function $h$ has no internal stationary points because the radial derivative is strictly positive:

$$r \frac{\partial h(x,y,z)}{\partial r} ~=~x \frac{\partial h(x,y,z)}{\partial x}+y \frac{\partial h(x,y,z)}{\partial y}+z \frac{\partial h(x,y,z)}{\partial z}$$ $$\tag{10}~=~\sum_{{\rm cycl.}~ x,y,z}\left(x^2y+2x^2-2xy\right) ~=~\sum_{{\rm cycl.}~ x,y,z}\left(x^2y+(x-y)^2\right)~>0.$$

Thus we may restrict to the sphere $x^2+y^2+z^2=3$ from now on. In other words, if we could prove that $$\tag{11} \forall x,y,z\geq 0:\quad x^2 +y^2 +z^2 = 3 \quad\Rightarrow \quad h(x,y,z)~\stackrel{?}{\geq}~ 0,$$

then eq. (9) would follow.

VII) Let us rewrite

$$\tag{12} h(x,y,z) ~=~ 2 + \underbrace{\left(-3+\sum_{{\rm cycl.}~ x,y,z}x^2\right)}_{= 0} - j(x,y,z),$$

where

$$\tag{13} j(x,y,z) ~:=~ \sum_{{\rm cycl.}~ x,y,z}k(x)y,$$

and where the concave function $k$ is

$$\tag{14} k(x) ~:=~ x(1-\frac{x}{3})$$

with derivative $k'(x) = 1-\frac{2x}{3}$. The tangent at $x=1$ is above the concave function,

$$\tag{15} k(x) ~\leq~ k(1) +(x-1) k'(1)~=~\frac{x+1}{3}.$$

If we could prove that

$$\tag{16} \forall x,y,z\geq 0:\quad x^2 +y^2 +z^2 = 3 \quad\Rightarrow \quad j(x,y,z)~\stackrel{?}{\leq}~ 2,$$

then eq. (11) would follow.

VIII) Proof of eq. (16):

$$\tag{17} j(x,y,z) ~\leq~\frac{1}{3}\sum_{{\rm cycl.}~ x,y,z}(x+1)y ~\leq~2.$$

The first inequality follows from eqs. (13) and (15). The second inequality follows because

$$\tag{18} \sum_{{\rm cycl.}~ x,y,z}xy ~\leq~ \sum_{{\rm cycl.}~ x,y,z}x^2 ~=~3,$$

and because $(x,y,z)=(1,1,1)$ is the maximum point for the function $(x,y,z)\mapsto x+y+z$ on the sphere $x^2+y^2+z^2=3$.

IX) Working backwards through the chain of arguments, we have proven OP's inequality.

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This is not the shortest proof, but at least it is elementary and correct. –  Qmechanic Jul 20 '12 at 10:08

Let $a=\frac{1}{x} ,b=\frac{1}{y} ,c=\frac{1}{z}$

$\frac{(x+1)(y+2)}{(y+1)(y+5)} + \frac{(y+1)(z+2)}{(z+1)(z+5)}+\frac{(z+1)(x+2)}{(x+1)(x+5)} \geq \frac{3}{2}$

note that $\frac{y+2}{(y+1)(y+5)}\ge \frac{3}{4(y+2)}$

then is enough to prove that $\frac{x+1}{y+2} + \frac{y+1}{z+2} + \frac{z+1}{x+2} \geq 2$

by Cauchy-Schwarz

$\frac{x+1}{y+2} + \frac{y+1}{z+2} + \frac{z+1}{x+2} =$

$\frac{(x+1)^2}{ (x+1)(y+2) }+\frac{(y+1)^2}{(y+1)(z+2)}+\frac{(z+1)^2}{(z+1)(x+2)}\geq \frac{((x+1)+(y+1)+(z+1))^2}{(x+1)(y+2) +(y+1)(z+2) +(z+1)(x+2) } \geq 2$

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+1, elegant and correct. –  Qmechanic Jul 20 '12 at 10:11