# I cannot prove that if f(x) is convex on [a,b]

I cannot prove that if $$\displaystyle{f{{\left({x}\right)}}}$$ is convex on $$\displaystyle{\left[{a},{b}\right]}$$ then
$$\displaystyle{f{{\left({\frac{{{a}+{b}}}{{{2}}}}\right)}}}\leq{\frac{{{1}}}{{{b}-{a}}}}{\int_{{a}}^{{b}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}\leq{\frac{{{f{{\left({a}\right)}}}+{f{{\left({b}\right)}}}}}{{{2}}}}$$

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otoplilp1
The inequality
$$\displaystyle{f{{\left({\frac{{{a}+{b}}}{{{2}}}}\right)}}}\leq{\frac{{{1}}}{{{b}-{a}}}}{\int_{{a}}^{{b}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$$
is a special case of Jensen's inequality.
And since f is convex, we have
$$\displaystyle{f{{\left({x}\right)}}}\leq{f{{\left({a}\right)}}}+{\frac{{{f{{\left({b}\right)}}}-{f{{\left({a}\right)}}}}}{{{b}-{a}}}}\cdot{\frac{{{\left({b}-{a}\right)}^{{2}}}}{{{2}}}}$$
$$\displaystyle={\left({b}-{a}\right)}{\frac{{{f{{\left({a}\right)}}}+{f{{\left({b}\right)}}}}}{{{2}}}}$$
which is equivalent to the second inequality.
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sukljama2
For convenience, we can take $$\displaystyle{b}={1},\ {a}=-{1}$$
For the first inequality, use
$$\displaystyle{f{{\left({0}\right)}}}\leq{\frac{{{f{{\left({x}\right)}}}+{f{{\left(-{x}\right)}}}}}{{{2}}}}$$
and integrate both sides from $$\displaystyle{x}=-{1}$$ to 1
For the second, use
$$\displaystyle{f{{\left({x}\right)}}}={f{{\left({\frac{{{1}-{x}}}{{{2}}}}{\left(-{1}\right)}+{\frac{{{1}+{x}}}{{{2}}}}{\left({1}\right)}\right)}}}\leq{\frac{{{1}-{x}}}{{{2}}}}{f{{\left(-{1}\right)}}}+{\frac{{{1}+{x}}}{{{2}}}}{f{{\left({1}\right)}}}$$
and integrate both sides from $$\displaystyle{x}=-{1}$$ to 1.
star233

Since f is convex ob [a,b], one has:
$$\forall x\in[a,b],\ f(x)\leq\frac{f(b)-f(a)}{b-a}(x-a)+f(a)$$
Therefore, one gets:
$$\int_a^b f(x)dx\leq\frac{f(b)-f(a)}{b-a}\frac{(b-a)^2}{2}+f(a)(b-a)$$
Finally, one has:
$$\frac{1}{b-a}\int_a^b f(x)dx\leq\frac{f(b)+f(a)}{2}$$