<s>[INST] Show that $\sqrt{ab} \leq \sqrt[n]{\frac{a^n+b^n+\lambda((a+b)^n-a^n-b^n)}{2+\lambda(2^n-2)}} \leq \frac{a+b}{2}$ Let $a,b,\lambda \in \mathbb R,$ with $a,b > 0, \lambda ≥1,$ and $ n \in \mathbb N^*.$
Show that $\sqrt{ab} \leq \sqrt[n]{\frac{a^n+b^n+\lambda((a+b)^n-a^n-b^n)}{2+\lambda(2^n-2)}} \leq \frac{a+b}{2}$
What is the good method to solve such a thing? I do not really have any idea for the moment ... It was given at an oral exam.
$(\sqrt{a}-\sqrt{b})^2 ≥0 \implies \sqrt{ab} \leq \frac{a+b}{2}$
 [/INST] I'll prove the right inequality
and leave the left.
$\sqrt{ab} \leq \sqrt[n]{\dfrac{a^n+b^n+\lambda((a+b)^n-a^n-b^n)}{2+\lambda(2^n-2)}} \leq \dfrac{a+b}{2}
$
Assume that
$a \le b$.
Dividing by $b$,
and writing $c = \lambda$
and
$x = a/b$,
this becomes
$\sqrt{a/b} 
\leq \sqrt[n]{\dfrac{(a/b)^n+1+c((a/b+1)^n-(a/b)^n-1)}{2+c(2^n-2)}} 
\leq \dfrac{(a/b)+1}{2}
$
or
$\sqrt{x} 
\leq \sqrt[n]{\dfrac{x^n+1+c((x+1)^n-x^n-1)}{2+c(2^n-2)}} 
\leq \dfrac{x+1}{2}
$
where
$0 < x < 1$
and
$c > 1$.
Let
$f(x)
=\dfrac{x^n+1+c((x+1)^n-x^n-1)}{2+c(2^n-2)}
$.
The right inequality is
$f(x)
\le \dfrac{(x+1)^n}{2^n}
$
or
$\begin{array}\\
x^n+1+c((x+1)^n-x^n-1)
&\le (2+c(2^n-2))(x+1)^n/2^n\\
&= c(x+1)^n+(2-2c)(x+1)^n/2^n\\
\text{or}\\
(1-c)x^n+1-c
&\le (2-2c)(x+1)^n/2^n\\
\text{or}\\
x^n+1
&\ge 2(x+1)^n/2^n
\qquad\text{since } c > 1\\
\text{or}\\
2^{n-1}(x^n+1)
&\ge (x+1)^n\\
\end{array}
$
Let
$g_n(x)
= 2^{n-1}(x^n+1)-(x+1)^n
$.
$g(0)=2^{n-1}-1 > 0$
and
$g(1) = 0$.
This is the same as
$\dfrac{x^n+1}{2}
\ge (\dfrac{x+1}{2})^n
$
or
$\sqrt[n]{\dfrac{x^n+1}{2}}
\ge \dfrac{x+1}{2}
$
which is true
by the power-mean inequality
(https://en.wikipedia.org/wiki/Generalized_mean).
That's enough for now.
The left inequality should be
similarly provable.
</s>