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f-I-0-I-R-f-twice-derivable-f-f-continuous-f-x-f-x-f-x-2-x-I-then-prove-that-2f-x-y-2-f-x-f-y-x-y-I-




Question Number 163209 by HongKing last updated on 04/Jan/22
f : I → (0 ; ∞)  ;  I ⊂ R  f - twice derivable  ;  f^′  ; f^(′′)  - continuous  f^(′′) (x) f(x) ≥ (f^′ (x))^2  ;  ∀ x ∈ I  then prove that:  2f (((x + y)/2)) ≤ f(x) + f(y)  ;  ∀ x;y ∈ I
$$\mathrm{f}\::\:\mathrm{I}\:\rightarrow\:\left(\mathrm{0}\:;\:\infty\right)\:\:;\:\:\mathrm{I}\:\subset\:\mathbb{R} \\ $$$$\mathrm{f}\:-\:\mathrm{twice}\:\mathrm{derivable}\:\:;\:\:\mathrm{f}\:^{'} \:;\:\mathrm{f}\:^{''} \:-\:\mathrm{continuous} \\ $$$$\mathrm{f}\:^{''} \left(\mathrm{x}\right)\:\mathrm{f}\left(\mathrm{x}\right)\:\geqslant\:\left(\mathrm{f}\:^{'} \left(\mathrm{x}\right)\right)^{\mathrm{2}} \:;\:\:\forall\:\mathrm{x}\:\in\:\mathrm{I} \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\mathrm{2f}\:\left(\frac{\mathrm{x}\:+\:\mathrm{y}}{\mathrm{2}}\right)\:\leqslant\:\mathrm{f}\left(\mathrm{x}\right)\:+\:\mathrm{f}\left(\mathrm{y}\right)\:\:;\:\:\forall\:\mathrm{x};\mathrm{y}\:\in\:\mathrm{I} \\ $$

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