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Question Number 143814 by mathdanisur last updated on 18/Jun/21

∀a;b;c∈R , find all f:R→R , such that f(a)f(bc)+9≤f(ab)+5f(ac)

$$\forall{a};{b};{c}\in\mathbb{R}\:,\:{find}\:{all}\:{f}:\mathbb{R}\rightarrow\mathbb{R}\:,\:{such}\:{that} \\ $$$${f}\left({a}\right){f}\left({bc}\right)+\mathrm{9}\leqslant{f}\left({ab}\right)+\mathrm{5}{f}\left({ac}\right) \\ $$

Answered by Olaf_Thorendsen last updated on 18/Jun/21

f(a)f(bc)+9 ≤ f(ab)+5f(ac) If a = b = c = 0 : f^2 (0)+9 ≤ 6f(0) f^2 (0)−6f(0)+9 ≤ 0 (f(0)−3)^2 ≤ 0 ⇒ f(0) = 3 If a = b = c = 1 : f(1) = 3 If a = 0 and c= 1 : f(0)f(b)+9 ≤ f(0)+5f(0) 3f(b)+9 ≤ 6f(0) = 18 f(b) ≤ 3 ∀b (1) If a = 1 and b = 0 : f(1)f(0)+9 ≤ f(0)+5f(c) 15 ≤ 5f(c) f(c) ≥ 3 ∀c (2) (1) and (2) ∀x, f(x) = 3

$${f}\left({a}\right){f}\left({bc}\right)+\mathrm{9}\:\leqslant\:{f}\left({ab}\right)+\mathrm{5}{f}\left({ac}\right) \\ $$$$ \\ $$$$\mathrm{If}\:{a}\:=\:{b}\:=\:{c}\:=\:\mathrm{0}\:: \\ $$$${f}^{\mathrm{2}} \left(\mathrm{0}\right)+\mathrm{9}\:\leqslant\:\mathrm{6}{f}\left(\mathrm{0}\right) \\ $$$${f}^{\mathrm{2}} \left(\mathrm{0}\right)−\mathrm{6}{f}\left(\mathrm{0}\right)+\mathrm{9}\:\leqslant\:\mathrm{0} \\ $$$$\left({f}\left(\mathrm{0}\right)−\mathrm{3}\right)^{\mathrm{2}} \:\leqslant\:\mathrm{0}\:\Rightarrow\:{f}\left(\mathrm{0}\right)\:=\:\mathrm{3} \\ $$$$ \\ $$$$\mathrm{If}\:{a}\:=\:{b}\:=\:{c}\:=\:\mathrm{1}\:: \\ $$$${f}\left(\mathrm{1}\right)\:=\:\mathrm{3} \\ $$$$ \\ $$$$\mathrm{If}\:{a}\:=\:\mathrm{0}\:\mathrm{and}\:\mathrm{c}=\:\mathrm{1}\:: \\ $$$${f}\left(\mathrm{0}\right){f}\left({b}\right)+\mathrm{9}\:\leqslant\:{f}\left(\mathrm{0}\right)+\mathrm{5}{f}\left(\mathrm{0}\right) \\ $$$$\mathrm{3}{f}\left({b}\right)+\mathrm{9}\:\leqslant\:\mathrm{6}{f}\left(\mathrm{0}\right)\:=\:\mathrm{18} \\ $$$${f}\left({b}\right)\:\leqslant\:\mathrm{3}\:\forall{b}\:\:\:\left(\mathrm{1}\right) \\ $$$$ \\ $$$$\mathrm{If}\:{a}\:=\:\mathrm{1}\:\mathrm{and}\:{b}\:=\:\mathrm{0}\:: \\ $$$${f}\left(\mathrm{1}\right){f}\left(\mathrm{0}\right)+\mathrm{9}\:\leqslant\:{f}\left(\mathrm{0}\right)+\mathrm{5}{f}\left({c}\right) \\ $$$$\mathrm{15}\:\leqslant\:\mathrm{5}{f}\left({c}\right) \\ $$$${f}\left({c}\right)\:\geqslant\:\mathrm{3}\:\forall{c}\:\:\:\left(\mathrm{2}\right) \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:\mathrm{and}\:\left(\mathrm{2}\right)\:\forall{x},\:{f}\left({x}\right)\:=\:\mathrm{3} \\ $$

Commented by mathdanisur last updated on 20/Jun/21

thankyou sir cool

$${thankyou}\:{sir}\:{cool} \\ $$

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