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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.

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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.$$ f(λ_1 x_1 + λ_2 x_2) \geqslant λ_1 f(x_1) + λ_2 f(x_2). $$

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  • $\begingroup$ Yes, that is the way to write $\lambda_1$ and $\lambda_2$ that I was looking for. Thank you. $\endgroup$ – MarianaMG2205 Mar 7 '19 at 2:24

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