Show that $\sqrt{ab} \leq \sqrt[n]{\frac{a^n+b^n+\lambda((a+b)^n-a^n-b^n)}{2+\lambda(2^n-2)}} \leq \frac{a+b}{2}$

Question

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.

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

Answer 1

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.

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

Answer 3

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.