Inequality

Problem : Let $x,y,m,n$ be positive real numbers which sastify $m\ne n$ and

$$x^m+y^m=x^n+y^n$$.

Prove that

$$x+y\le 2$$.

My Solution :

Firstly, we will prove that :

If $x,y$ be positive real numbers and $m,n$ be positive rational numbers such that $x^m+y^m=x^n+y^n$, we get $x+y\le 2$.

Let $m=\frac{A}{B},n=\frac{C}{D}$ with $A,B,C,D$ are positive integers and $\gcd (A,B)=1,\gcd (C,D)=1$.

Wlog, we can suppose $m>n$, then $DA>BC$.

By Holder Inequality :

$\underset{BCtimes}{\underbrace{\left(x^{\frac{A}{B}}+y^{\frac{A}{B}}\right)...\left(x^{\frac{A}{B}}+y^{\frac{A}{B}}\right)}}\underset{DA-BCtimes}{\underbrace{(1+1)...(1+1)}}$$\ge {\left(\sqrt[DA]{{\left(x^{\frac{A}{B}}\right)}^{BC}}+\sqrt[DA]{{\left(y^{\frac{A}{B}}\right)}^{BC}}\right)}^{DA}$

 $\iff {\left(x^{\frac{A}{B}}+y^{\frac{A}{B}}\right)}^{BC}.2^{DA-BC}$$\ge {\left(x^{\frac{C}{D}}+y^{\frac{C}{D}}\right)}^{DA-BC}$

It’s easy for us to get :

$$x^{\frac{A}{B}}+y^{\frac{A}{B}}\le 2$$

with the assumption of this problem $x^{\frac{A}{B}}+y^{\frac{A}{B}}=x^{\frac{C}{D}}+y^{\frac{C}{D}}$.

Applying Holder Inequality again :

$\underset{Btimes}{\underbrace{\left(x^{\frac{A}{B}}+y^{\frac{A}{B}}\right)...\left(x^{\frac{A}{B}}+y^{\frac{A}{B}}\right)}}\underset{A-Btimes}{\underbrace{(1+1)...(1+1)}}$$\ge {\left(\sqrt[A]{{\left(x^{\frac{A}{B}}\right)}^B}+\sqrt[A]{{\left(y^{\frac{A}{B}}\right)}^B}\right)}^A$

$$\iff {\left(x^{\frac{A}{B}}+y^{\frac{A}{B}}\right)}^B.2^{A-B}\ge (x+y{)}^A$$

But we got $x^{\frac{A}{B}}+y^{\frac{A}{B}}\le 2$, so $x+y\le 2$.

For all the real number $m,n$, there always exist two rational sequences $(a_n),(b_n)$ sastifying that :

$$\underset{k\to +\infty }{\lim }{a}_k=m,\underset{k\to +\infty }{\lim }{b}_k=n$$

As what we proved above :

If $$x^m+y^m=x^n+y^n$$ with $$x,y\in {\mathbb{R}}^{+}$$ for all $$m,n\in {\mathbb{Q}}^{+}$$, we have $$x+y\le 2$$.

As a result, we get :

If $$x^{a_k}+y^{a_k}=x^{b_k}+y^{b_k},\forall k\in {\mathbb{N}}^{\ast }$$, then $$x+y\le 2$$.

Let $k\to +\infty$, it’s obivious that :

If $$x^{\lim a_k}+y^{\lim a_k}=x^{\lim b_k}+y^{\lim b_k}$$, then $$x+y\le 2$$

Or :

If $$x^m+y^m=x^n+y^n$$, then $$x+y\le 2$$.

This is all what we’re in need of proving.