How to prove$\sum\frac{a}{a^2+a+1}\leq\sum\frac{1}{a^2+a+1}$

Question

If I asked the obvious because of my lack of knowledge, then I am sorry!

If $abcd=1,a,b,c,d>0$;prove:$$\sum_\limits{cyc}\frac{a}{a^2+a+1}\leq\sum_\limits{cyc}\frac{1}{a^2+a+1}$$

I use $f(x)=\frac{x-1}{x^2+x+1}$,it without success;I want to use$a=\frac{yz}{x^2},b=\frac{zt}{y^2},c=\frac{tx}{z^2},d=\frac{xy}{t^2},$and the Order inequality,but I'm not sure if it's right.Could anyone help me? Thanks a lot!

Answer 1

$\sum\limits_{cyc}\frac{1-a}{1+a+a^2}=\sum\limits_{cyc}\left(\frac{1-a}{1+a+a^2}+\frac{1}{3}\ln{a}\right)$.

Let $f(x)=\frac{1-x}{x^2+x+1}+\frac{1}{3}\ln{x}$.

Hence, $f'(x)=\frac{(x-1)(x^3+6x^2+3x-1)}{3x(x^2+x+1)^2}$.

Which says that there is unique $0<x_1<1$, for which $f(x)\geq0$ for all $x\geq x_1$.

Easy to see that $x_1=0.0779...$,

which says that for $\min\{a,b,c,d\}\geq0.08$ our inequality is true.

Let $a<0.08$ and $g(x)=\frac{1-x}{1+x+x^2}$.

Hence, $g'(x)=\frac{x^2-2x-2}{(1+x+x^2)^2}$, which says that $\min\limits_{x>0}g=g(1+\sqrt3)$

and $g$ is a decreasing function on $(0,0.08]$.

Id est, $\sum\limits_{cyc}\frac{1-a}{1+a+a^2}\geq g(0.08)+3g(1+\sqrt3)=0.38...>0$

Done!

Answer 2

If $abcd =1$, then put $f(x,y,z,w) = x+y+z+w$ subject to the constraint $g(x,y,z,w) = xyzw -1 =0$. Basically, we want to show that $a = b = c= d$ is a maximum when they all equal $1$.

So, we have, using lagrange multipliers,

• $f_x = 1 = \lambda g_x = \lambda yzw$
• $f_y = 1 = \lambda g_y = \lambda xzw$
• $f_z = 1 = \lambda g_z = \lambda xyw$
• $f_w = 1 = \lambda g_w = \lambda xyz$
• $g(x,y,z,w) = xyzw -1 = 0$.

So, we have $xyz = \frac{1}{\lambda},$ hence $\frac{w}{\lambda} =1 \Rightarrow w = \lambda$.

But then $yz = \frac{1}{\lambda w} = \frac{1}{\lambda^2}$ so $x = \lambda$. Continuing in this fashion, we get $x = y =z = w =\lambda$, so $\lambda^4 = 1$. Clearly $\lambda =1$, rather than $\lambda = -1$, gives us our max for $f$.

Putting it all together, we have $\displaystyle \sum_{cyc} \frac{1-a}{a^2 +a +1} \ge 0$ for all $a,b,c,d$ such that $abcd =1$. This completes the proof.