Prove $f(x)=(x-3)^2 +4$ is a convex function

Question

for convex functions: $$\lambda f(x_1)+(1-\lambda)f(x_2)\geq f(\lambda x_1 +(1-\lambda)x_2)$$ for $\lambda \in [0, 1]$ My solution looks like: $$\lambda (x_1-3)^2 +4\lambda +(1-\lambda)((x-3)^2 +4)\ge\bigl(\lambda x_1+(1-\lambda)x_2 - 3)^2+4$$ $$\lambda (x_1^2+9-6x_1)+4\lambda+(1-\lambda)(x_2^2+13-6x_2) \ge (\lambda x_1+(1-\lambda)x_2-3)^2 +4 $$ $$\lambda x_1^2 +x_2^2-\lambda x_2^2-6\lambda x_1-6\lambda x_2+13 \ge \lambda^2x_1^2+x_2^2+\lambda^2x_2^2-\lambda x_2^2+9+2\lambda x_1-2\lambda^2 x_2 x_1-6x_2+6\lambda x_2-6\lambda x_1$$ $$0 \ge\lambda x_1^2 - \lambda^2x_1^2 +\lambda^2x_2^2 +2\lambda x_1+12\lambda x_2 -2\lambda^2x_2x_1-4 $$

I am not sure if above solution is correct, but I could not figure out how to proceed from this point.

Answer 1

We need to show that $$\lambda (x_1-3)^2 +4\lambda +(1-\lambda)(x_2-3)^2 +4-4\lambda \ge\bigl(\lambda x_1+(1-\lambda)x_2 - 3)^2+4.$$ That is,

$$\lambda (x_1-3)^2 +(1-\lambda)(x_2-3)^2 \ge\bigl(\lambda x_1+(1-\lambda)x_2 - 3)^2.$$ Equivalently $$\lambda x_1^2+(1-\lambda)x_2^2-6\lambda x_1- 6(1-\lambda) x_2+9$$ $$\ge \lambda^2 x_1^2+(1-\lambda)^2 x_2^2+2\lambda(1-\lambda)x_1x_2-6\lambda x_1-6(1-\lambda)x_ 2+9$$

Simplifying, we arrive at

$$\lambda x_1^2+(1-\lambda)x_2^2+\ge \lambda^2 x_1^2+(1-\lambda)^2 x_2^2+2\lambda(1-\lambda)x_1x_2.$$ Arranging terms

$$\lambda(1-\lambda) x_1^2+\lambda(1-\lambda)x_2^2-2\lambda(1-\lambda)x_1x_2\ge 0.$$ This inequality is equivalent to

$$ \lambda(1-\lambda)(x_1-x_2)^2=\lambda(1-\lambda)(x_1^2+x_2^2-2x_1x_2)\ge 0,$$ which clearly holds.

Answer 2

HINT

Use the second derivative

Answer 3

Note that we can write

$$\begin{align} f(\lambda x_1+(1-\lambda) x_2)&=f(\lambda (x_1-3)+(1-\lambda )(x_2-3))\\\\ &=(\lambda (x_1-3)+(1-\lambda)(x_2-3))^2+4\\\\ &=(\lambda (x_1-3))^2+((1-\lambda)(x_2-3))^2\\\\ &+2\lambda (1-\lambda)(x_1-3)(x_2-3)+4 \tag 1 \end{align}$$

Furthermore, we can write

$$\lambda f(x_1)+(1-\lambda)f(x_2)=\lambda (x_1-3)^2+(1-\lambda)(x_2-3)^2+4 \tag 2$$

Subtracting $(1)$ from $(2)$ yields

$$\begin{align} \lambda f(x_1)+(1-\lambda)f(x_2)-f(\lambda x_1+(1-\lambda) x_2)&=\lambda (1-\lambda)(x_1-x_2)^2\\\\ &\ge 0 \end{align}$$

as was to be shown!