# Proving an inequality regarding convex functions

Let $$f$$ be a real-valued convex function, $$\lambda_1>0$$ and $$\lambda_2\leq0$$ such that $$\lambda_1+\lambda_2=1$$.

I want to prove that for any $$x_1$$,$$x_2\in{\rm Dom}(f)$$, $$f(\lambda_1x_1+\lambda_2x_2)\geq \lambda_1f(x_1)+\lambda_2f(x_2).$$

Now, since $$f$$ is convex, for any $$t\in[0,1]$$, $$f(tx_1+(1-t)x_2)\leq tf(x_1)+(1-t)f(x_2).$$ I noticed that the inequality that I want to prove is very similar to the definition of convexity, but taking values outside the interval $$[0,1]$$ that sum to $$1$$. I know that convexity means that the graph of f between $$x_1$$ and $$x_2$$ is below the segment which joins the points $$(x_1,f(x_1))$$ and $$(x_2,f(x_2))$$, so maybe there is some geometric interpretation for $$\lambda_1x_1+\lambda_2x_2$$.

It feels that with some inequalities I should be solve this problem, but as of now, I cannot see them. Any help is appreciated.

Note that $$\dfrac{1}{λ_1} + \dfrac{-λ_2}{λ_1} = 1$$ and $$λ_1 > 0 \geqslant λ_2$$, thus$$\frac{1}{λ_1} f(λ_1 x_1 + λ_2 x_2) + \dfrac{-λ_2}{λ_1} f(x_2) \geqslant f\left( \frac{1}{λ_1} · (λ_1 x_1 + λ_2 x_2) + \left( \frac{-λ_2}{λ_1} \right) · x_2 \right) = f(x_1),$$ i.e.$$f(λ_1 x_1 + λ_2 x_2) \geqslant λ_1 f(x_1) + λ_2 f(x_2).$$
• Yes, that is the way to write $\lambda_1$ and $\lambda_2$ that I was looking for. Thank you. – MarianaMG2205 Mar 7 '19 at 2:24