How to prove $f(x)=x^4$ is concave up by definition?

Question

I know $f(x)=x^4$ is concave up, by calculating its second derivative. However, how to prove that $f(x)=x^4$ is concave up by definition, say $f(\frac{x_1+x_2}{2})<(1/2)f(x_1)+(1/2)f(x_2)$ for all $x_1,~x_2$? I'd tried binomial theorem, but can't get anything.

Answer 1

As an intermediate step, use the inequality

$$ \frac{x^2 + y^2}{2} \ge \left(\frac{x + y}{2}\right)^2 $$

which is relatively easy to prove.

This gives

$$ \frac{x^4 + y^4}{2} = \frac{(x^2)^2 + (y^2)^2}{2} \ge \left(\frac{x^2 + y^2}{2}\right)^2 \ge \left(\left(\frac{x + y}{2}\right)^2\right)^2 = \left(\frac{x + y}{2}\right)^4. $$

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Alternatively, using the "usual" definition of convexity, we have

$$ [(1 - \lambda)x^2 + \lambda y^2] - [(1 - \lambda)x + \lambda y]^2 = (1 - \lambda)\lambda (x - y)^2 \ge 0. $$

Being a quadratic form, this is an easy enough sum of squares to compute. One can do it by hand even. Then we do what we did above:

$$ (1 - \lambda)x^4 + \lambda y^4 \ge [(1 - \lambda)x^2 + \lambda y^2]^2 \ge [(1 - \lambda)x + \lambda y]^4 . $$

Answer 2

I think you mean that $f$ is a convex function.

Id est, we need to prove that: $$\frac{a^4+b^4}{2}\geq\left(\frac{a+b}{2}\right)^4$$ or $$8(a^4+b^4)\geq a^4+4a^3b+6a^2b^2+4ab^3+b^4$$ or $$7a^4-4a^3b-6a^2b^2-4ab^3+7b^4\geq0$$ or $$7a^4-14a^3b+7a^2b^2+10a^3b-20a^2b^2+10ab^3+7a^2b^2-14ab^3+7b^4\geq0$$ or $$(a-b)^2(7a^2+10ab+7b^2)\geq0$$ or $$(a-b)^2(2a^2+2b^2+5(a+b)^2)\geq0,$$ which is obvious.