A property of Nesbitt's inequality

Question

I think it's not new however I think it's interesting :

Let $a,b,c>0$ and $x>0$ then the function : $$f(x)=\frac{a^x}{b^x+c^x}+\frac{b^x}{a^x+c^x}+\frac{c^x}{b^x+a^x}$$ $f(x)$ is increasing

My first try :

The derivative of $f(x)$ is :

$$f'(x)=\sum_{cyc}\frac{a^x\ln(a)(b^x+c^x)-a^x(b^x\ln(b)+c^x\ln(c))}{(b^x+c^x)^2}$$

We multiply by $x$ it gives : $$f'(x)=\frac{1}{x}\Big(\sum_{cyc}\frac{a^x\ln(a^x)(b^x+c^x)-a^x(b^x\ln(b^x)+c^x\ln(c^x))}{(b^x+c^x)^2}\Big)$$

$x$ is positive so we have to prove :

$$\sum_{cyc}\frac{a^x\ln(a^x)(b^x+c^x)-a^x(b^x\ln(b^x)+c^x\ln(c^x))}{(b^x+c^x)^2}\geq 0$$

We put $a^x=u$ , $b^x=v$ and $c^x=w$ so the inequality is equivalent to : $$\sum_{cyc}\frac{u\ln(u)(v+w)-u(v\ln(v)+w\ln(w))}{(v+w)^2}\geq 0$$

Now with the concavity of $\ln(x)$ and the inequality of Jensen's we have :

$$\ln\Big(\frac{v^2+w^2}{v+w}\Big)\geq \frac{v\ln(v)+w\ln(w)}{v+w}$$

So we have :

$$\sum_{cyc}\frac{u\ln(u)(v+w)-u(v\ln(v)+w\ln(w))}{(v+w)^2}\geq\sum_{cyc}\frac{u\ln(u)-u\Big(\ln\Big(\frac{v^2+w^2}{v+w}\Big)\Big)}{(v+w)}\geq 0$$

Or :

$$\sum_{cyc}\frac{u\ln(u)(v+w)-u(v\ln(v)+w\ln(w))}{(v+w)^2}\geq\sum_{cyc}\frac{u\Big(\ln\Big(\frac{u(v+w)}{v^2+w^2}\Big)\Big)}{(v+w)}\geq 0$$

Now we have to prove :

$$\sum_{cyc}\frac{u\Big(\ln\Big(\frac{u(v+w)}{v^2+w^2}\Big)\Big)}{(v+w)}\geq 0$$

Unfortunatly I'm stuck here ...

Second try :

For $a,b,c>0$ the function $g(x)=\frac{a^x}{b^x+c^x}$ is convex so the function $f(x)$ is convex as sum of convex functions .

Now we have with $a,b,c>0$ :

$$\sum_{cyc}\frac{a^2}{b^2+c^2}\geq \sum_{cyc}\frac{a}{b+c}$$

Fact wich have been proven by Michael Rozenberg (see my questions)

Remains to apply the three chord lemma to get the increase of the function $f(x)$

My question :

Can someone could find an alternative proof or prove this last inequality ?

There are generalizations of this facts ?

Answer 1

By your assumption $$f'(x)=\sum_{cyc}\frac{a^x\ln{a}(b^x+c^x)-a^x(b^x\ln{b}+c^x\ln{c})}{(b^x+c^x)^2}=$$ $$=\sum_{cyc}\frac{a^x(b^x(\ln{a}-\ln{b})-c^x(\ln{c}-\ln{a}))}{(b^x+c^x)^2}=$$ $$=\sum_{cyc}(\ln{a}-\ln{b})\left(\frac{a^xb^x}{(b^x+c^2)^2}-\frac{a^xb^x}{(c^x+a^x)^2}\right)=$$ $$=\sum_{cyc}\frac{a^xb^x(\ln{a}-\ln{b})(a^x-b^x)(a^x+b^x+2c^x)}{(b^x+c^x)^2(c^x+a^x)^2}\geq0$$ because $$(\ln{a}-\ln{b})(a^x-b^x)=b^x\ln\frac{a}{b}\left(\left(\frac{a}{b}\right)^x-1\right)\geq0$$ for all $x>0$ and we are done!

Answer 2

Lemma: [Smoothing of convex functions] Given a convex function $f(x)$, and variables $ a \leq b$, if $ \epsilon < b-a$, then $ f(a) + f(b) \geq f( a + \epsilon ) + f(b- \epsilon)$.

(This is "well-known", so I'm not going to prove it. If you're stuck, explain what you've tried.)

Corollary: Given a convex function $f(x)$, and variables $ a \leq b \leq c$, $ A \leq B \leq C$ such that $ a + b + c = A + B + C $, $ C \geq c$, and $A \leq a$, then $ f(A) + f(B) + f(C) \geq f(a) + f(b) + f(c)$.

Comments:
The proof follows by applying the above lemma (or see the hidden block).
I consider this "less well-known", though it's still often used for Olympiad inequalities.
The 4-variable analogue need not hold, because we don't have enough control over the middle terms.
I (personally) call this the "3-variable smoothing Jensens".

Proof: Let $ C - c = c' $, $ A - a = - a' $, $B - b = a' - c' $.
If $B - b > 0$, then $ f(B) + f(A) \geq f(b) + f(A+B-b)$ by the lemma.
We now have $ C + (A+B-b) = c + a$, so $ f(C) + f(A+B-b) \geq f(c) + f(a) $ by the lemma.
Thus, $ f(A) + f(B) + f(C) \geq f(a) + f(b) + f(c)$.
Likewise if $ B - b < 0$, then work with $ f(B) + f(C) \geq f(b) + f(B+C - b)$ first.

Corollary: The problem follows.

First, a simplification: To show that the function is increasing, we just need to show that for $ z = x/y > 1$, $$\sum \frac{ \frac{ p^z } {p^z + q^z + r^z } } { 1 - \frac{ p^z } {p^z + q^z + r^z } } \geq \frac{\frac{ p } { p + q + r }}{1 - \frac{ p } { p + q + r } } \quad (1) $$
then apply the substitution $ p = a^x, q = b^x, r = c^x$ to conclude that $f(y) \geq f(x)$.

Now, WLOG $ p = \max (p, q, r)$.
$ \frac{ p^z } {p^z + q^z + r^z } \geq \frac{ p } { p + q + r } \Leftrightarrow p^z q + p^z r \geq p q^z + p r^z$, which is true.
Likewise, WLOG $ r = \min (p,q,r)$ and $ \frac{ r^z } {p^z + q^z + r^z } \leq \frac{ r } { p + q + r }$

Hence, we can apply the previous corollary with $ f(x) = \frac{x}{1-x}$, $\{A, B, C \} = \{ \frac{ p^z } {p^z + q^z + r^z }, \frac{ q^z } {p^z + q^z + r^z }, \frac{ r^z } {p^z + q^z + r^z } \} $, $ \{ a, b, c \} = \{ \frac{ p } { p + q + r }, \frac{ q } { p + q + r }, \frac{ r } { p + q + r }\}$, so inequality (1) is true.

Answer 3

Alternative solution:

WLOG, assume $a \ge b \ge c = 1$. Denote $a^x = u, b^x = v$. Then $u \ge v \ge 1$.

We have \begin{align*} f'(x) &= A \ln a + B\ln b \\ &= A\ln \frac{a}{b} + (A + B) \ln b \end{align*} where \begin{align*} A &= {\frac {u}{v+1}} - {\frac {vu}{ \left( 1+u \right) ^{2}}} -{\frac {u}{ \left( u+v \right) ^{2}}},\\ B &={\frac {v}{1+u} -{\frac {vu}{ \left( v+1 \right) ^{2}}}} -{\frac {v}{ \left( u+v \right) ^{2}}}. \end{align*} We have $$A \ge {\frac {u}{v+1}} - {\frac {vu}{ \left( 1+v \right) ^{2}}} -{\frac {u}{ \left( 1+v \right) ^{2}}} = 0.$$ Also, we have $$A + B = \frac{u}{(1 + v)^2} - \frac{u}{(u + v)^2} + \frac{v}{(1 + u)^2} - \frac{v}{(u + v)^2} \ge 0. $$ Thus, $f'(x) \ge 0$ for all $x > 0$.

We are done.