How to show $ \left(\frac{1-x}{2}\right)^p+\left(\frac{1+x}{2}\right)^p \leq \frac{1+x^p}{2}$

Question

When $p\geq 2$ and $0\leq x\leq1$, how does one show the inequalities $$ \left(\frac{1-x}{2}\right)^p+\left(\frac{1+x}{2}\right)^p \leq \frac{1+x^p}{2}$$ and $$ 2(1+x^p)\leq (1+x)^p + (1-x)^p \ ?$$

The first one looks like a correction of the parallelogram law for powers greater than $p=2$, I've tried using that and rearranging but it hasn't worked. I've also tried differentiating but then I don't know how to handle the new expressions.

This is from a qualification exam on analysis. Basically the problem becomes trivial if one shows the inequalities.

Thank you!

This looks like Clarkson's identities – Surb Aug 12 '14 at 19:38 — Surb, Aug 12 '14 at 19:38

score 2 · Accepted Answer · answered Aug 12 '14 at 19:43

For the second one, the left hand side decreases as the value of P increases and the right hand side increases as the value of P increases. So if you show the smallest case is true(P=2 and x should be chosen to maximize the left hadn side) then I think that would be prove the second inequality. Of course you would also need to prove the LHS decreases and the RHS increases as P increases.

How to show $ \left(\frac{1-x}{2}\right)^p+\left(\frac{1+x}{2}\right)^p \leq \frac{1+x^p}{2}$

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