Show that $a^{2014}+b^{2014}\geq a^{2013}+b^{2013} $.

Question

Let $ a, b\in \mathbb {R}_{+} $ s.t. $a^{22}+b^{22}=a^{3}+b^{3} $. Show that $$a^{2014}+b^{2014}\geq a^{2013}+b^{2013} $$

By Chebyshev's inequality we obtain $$a^{19}+b^{19}\leq 2\Leftrightarrow b^{19}-1\leq 1-a^{19}\Leftrightarrow $$ $$ (b-1)(b^{18}+b^{17}+...+b+1)\leq (1-a)(a^{18}+a^{17}+...+a+1) $$

I supposed $b\geq 1\geq a $. Then $ b^{18}+b^{17}+...+b+1\geq a^{18}+a^{17}+...+a+1$. Then $b-1\leq 1-a \Leftrightarrow a+b\leq 2$. Now I am stuck.

Answer 1

Suppose $a,\ b\ne 1$. Let $f(x)=a^x+b^x$. The function is continuous, so its derivative has a root in $[3,\ 22]$. For $x_0$ to be a root of the derivative, the equation $$\left(\frac ab\right)^{x_0}=-\frac{\ln{b}}{\ln{a}}$$ must hold. We can see that $x_0$ is unique, so the only root of $f'$ is in $[3,\ 22]$. Since for $x\rightarrow\infty$ we have $f(x)\rightarrow\infty$ (at least one of $a,\ b$ is $>1$), the derivative is positive for $x>22$, so $f(m)>f(n)$ for all $m>n>22$. In the case $m=2014$ and $n=2013$, we get the above inequality.

Answer 2

Let $f(x)=a^x+b^x\ \ \ \ , \ \ \ \ \ f(3)=f(22)$

By Holder inequality:

$$ \left( f(3) \right )^{\frac{1991}{2010}}\cdot (f(2013))^{1-\frac{1991}{2010}} \ge f(22) \Rightarrow f(2013)\ge f(22)=f(3)$$

$$\left( f(22 \right )^{\frac{1}{1992}}\cdot (f(2014))^{1-\frac{1}{1992}} \ge f(2013) \Rightarrow f(2014) \ge f(2013)$$

Answer 3

We need to prove that $$\sum_{cyc}\left(a^{2014}-a^{2013}\right)\geq0$$ or $$\sum_{cyc}\left(a^{2014}-a^{2013}-\frac{1}{19}\left(a^{22}-a^3\right)\right)\geq0$$ or $$\sum_{cyc}a^3\left(19a^{2011}-19a^{2010}-a^{19}+1\right)\geq0.$$ Now, by AM-GM $$2010a^{2011}-2011a^{2010}+1\geq0$$ or $$19a^{2011}-\frac{19\cdot2011}{2010}a^{2010}+\frac{19}{2010}\geq0.$$ Id est, it's enough to prove that $$\left(\frac{19\cdot2011}{2010}-19\right)a^{2010}-a^{19}+1-\frac{19}{2010}\geq0$$ or $$19a^{2010}-2010a^{19}+1991\geq0,$$ which is AM-GM again.

Done!