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.
$(\sqrt{a}-\sqrt{b})^2 ≥0 \implies \sqrt{ab} \leq \frac{a+b}{2}$
 A: At $\lambda=1$ we have equality between the middle term and the RHS. If you manage to prove that the middle term is a decreasing function of $\lambda$, it just remains to show that
$$ (ab)^{n/2}\leq \frac{(a+b)^n-a^n-b^n}{2^n-2}. $$
This is a homogeneous inequality, hence it is enough to show that
$$ (2^n-2) x^{n/2} \leq (1+x)^n-1-x^n $$
for any $x\geq 0$. This is a simple consequence of the binomial theorem and AM-GM.
A: the right-hand side of your inequality is equivalent to $$\frac{a^n+b^n}{2}\geq \left(\frac{a+b}{2}\right)^n$$
ok, lets prove it,raise to the power $n$ gives:
$$a^n+b^n+\lambda(a+b)^n-a^n\lambda-b^n\lambda\le \left(\frac{a+b}{2}\right)^n(2+2^n\lambda-2\lambda)$$
after simplifying and rearranging we obtain:
$$(a^n+b^n)(1-\lambda)\le \left(\frac{a+b}{2}\right)^n2+(a+b)^n+(a+b)^n\lambda-2\lambda\left(\frac{a+b}{2}\right)^n$$
and so we get
$$(a^n+b^n)(1-\lambda)\le 2(1-\lambda)\left(\frac{a+b}{2}\right)^n$$
for $\lambda=1$ is the inequality true, for $$\lambda>1$$ we have
$$\frac{a^n+b^n}{2}\geq \left(\frac{a+b}{2}\right)^n$$
A: 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.
