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.
$(\sqrt{a}-\sqrt{b})^2 ≥0 \implies \sqrt{ab} \leq \frac{a+b}{2}$

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
$$\frac{a^n+b^n}{2}\geq \left(\frac{a+b}{2}\right)^n$$
