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I have to prove that the sum of convex functions is again convex. I know the definition of convex function: $f(tx_1+(1-t)x_2)\leq t f(x_1)+(1-t)f(x_2)$ - this the first convex function, then I have the second one $g(tx_1+(1-t)x_2)\leq t g(x_1)+(1-t)g(x_2)$

What should I do next? Thank you for your help and time.

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Hint: $$(f+g)(tx_1 +(1-t)x_2)=f(tx_1 +(1-t)x_2)+g(tx_1 +(1-t)x_2)\leq ...$$

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Try adding the two inequalities.

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Using the alternative definition of convexity can make the algebra somewhat more intuitive.

Definition of convexity

Let $f$ be a real function defined on a real interval $I$, $f$ is convex on $I$ if and only if:

$$ f(\alpha x + \beta y) \leq \alpha f(x) + \beta f(y)$$

$$ \forall x,y \in I \ : \forall \alpha, \beta \in \mathbb{R}_{\geq 0}, \ \alpha + \beta = 1 \ $$

First steps of the proof

let $h(x) = f(x) + g(x)$ where $f(x)$ and $g(x)$ are both convex functions...

I suppose we don't want to ruin the fun, but a quick google search gives a zillion versions of this proof.

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