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Question Number 219606 by Nicholas666 last updated on 29/Apr/25
prove that for positive real numbers a,b,c, the following inequality holds; (a^2 /(b + c)) + (b^2 /(c + a)) + (c^2 /(a + b)) ≥ ((a + b + c)/2)
$$ \\ $$$$\:\mathrm{prove}\:\mathrm{that}\:\mathrm{for}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{numbers}\:{a},{b},{c},\:\:\: \\ $$$$\mathrm{the}\:\mathrm{following}\:\mathrm{inequality}\:\mathrm{holds}; \\ $$$$\:\:\frac{{a}^{\mathrm{2}} }{{b}\:+\:{c}}\:+\:\frac{{b}^{\mathrm{2}} }{{c}\:+\:{a}}\:+\:\frac{{c}^{\mathrm{2}} }{{a}\:+\:{b}}\:\:\geqslant\:\frac{{a}\:+\:{b}\:+\:{c}}{\mathrm{2}} \\ $$$$ \\ $$
Answered by Nicholas666 last updated on 29/Apr/25
a,b,c >0 (a^2 /(b+c))+a = ((a(a+b+c))/(b+c)) (b^2 /(c+a))+b = ((b(a+b+c))/(c+a)) (c^2 /(a+b))+c = ((c(a+b+c))/(a+b)) ((a^2 /(b+c)) + (b^2 /(c+a)) + (c^2 /(a+b)))+(a+b+c)=(a+b+c)((a/(b+c))+(b/(c+a))+(c/(a+b))) (a/(b+c))+1=((a+b+c)/(b+c)) (b/(c+a))+1=((a+b+c)/(c+a)) (c/(a+b))+1=((a+b+c)/(a+b)) ((a/(b+c))+(b/(c+a))+(c/(a+b)))+3=(a+b+c)((1/(b+c))+(1/(c+a))+(1/(a+b))) (1/(b+c))+(1/(c+a))+(1/(a+b)) ≥ (9/(2(a+b+c))) (from AM−HM) ((a/(b+c))+(b/(c+a))+(c/(a+b)) ≥(3/2) ( (a^2 /(b+c))+(b^2 /(c+a))+(c^2 /(a+b)))+(a+b+c)≥(a+b+c).(3/2) (a^2 /(b+c))+(b^2 /(c+a))+(c^2 /(a+b)) ≥(1/2)(a+b+c) determinant ((((a^2 /(b+c)) + (b^2 /(c+a)) + (c^2 /(a+b)) ≥ ((a+b+c)/2) ✓)),(( the inequality is proven True)))
$${a},{b},{c}\:>\mathrm{0} \\ $$$$\frac{{a}^{\mathrm{2}} }{{b}+{c}}+{a}\:=\:\frac{{a}\left({a}+{b}+{c}\right)}{{b}+{c}} \\ $$$$\frac{{b}^{\mathrm{2}} }{{c}+{a}}+{b}\:=\:\frac{{b}\left({a}+{b}+{c}\right)}{{c}+{a}} \\ $$$$\frac{{c}^{\mathrm{2}} }{{a}+{b}}+{c}\:=\:\frac{{c}\left({a}+{b}+{c}\right)}{{a}+{b}} \\ $$$$\left(\frac{{a}^{\mathrm{2}} }{{b}+{c}}\:+\:\frac{{b}^{\mathrm{2}} }{{c}+{a}}\:+\:\frac{{c}^{\mathrm{2}} }{{a}+{b}}\right)+\left({a}+{b}+{c}\right)=\left({a}+{b}+{c}\right)\left(\frac{{a}}{{b}+{c}}+\frac{{b}}{{c}+{a}}+\frac{{c}}{{a}+{b}}\right) \\ $$$$\frac{{a}}{{b}+{c}}+\mathrm{1}=\frac{{a}+{b}+{c}}{{b}+{c}} \\ $$$$\frac{{b}}{{c}+{a}}+\mathrm{1}=\frac{{a}+{b}+{c}}{{c}+{a}} \\ $$$$\frac{{c}}{{a}+{b}}+\mathrm{1}=\frac{{a}+{b}+{c}}{{a}+{b}} \\ $$$$\left(\frac{{a}}{{b}+{c}}+\frac{{b}}{{c}+{a}}+\frac{{c}}{{a}+{b}}\right)+\mathrm{3}=\left({a}+{b}+{c}\right)\left(\frac{\mathrm{1}}{{b}+{c}}+\frac{\mathrm{1}}{{c}+{a}}+\frac{\mathrm{1}}{{a}+{b}}\right) \\ $$$$\frac{\mathrm{1}}{{b}+{c}}+\frac{\mathrm{1}}{{c}+{a}}+\frac{\mathrm{1}}{{a}+{b}}\:\geqslant\:\frac{\mathrm{9}}{\mathrm{2}\left({a}+{b}+{c}\right)}\:\:\:\:\:\:\:\left({from}\:\:{AM}−{HM}\right) \\ $$$$\left(\frac{{a}}{{b}+{c}}+\frac{{b}}{{c}+{a}}+\frac{{c}}{{a}+{b}}\:\geqslant\frac{\mathrm{3}}{\mathrm{2}}\right. \\ $$$$\left(\:\frac{{a}^{\mathrm{2}} }{{b}+{c}}+\frac{{b}^{\mathrm{2}} }{{c}+{a}}+\frac{{c}^{\mathrm{2}} }{{a}+{b}}\right)+\left({a}+{b}+{c}\right)\geqslant\left({a}+{b}+{c}\right).\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\frac{{a}^{\mathrm{2}} }{{b}+{c}}+\frac{{b}^{\mathrm{2}} }{{c}+{a}}+\frac{{c}^{\mathrm{2}} }{{a}+{b}}\:\geqslant\frac{\mathrm{1}}{\mathrm{2}}\left({a}+{b}+{c}\right) \\ $$$$\begin{array}{|c|c|}{\frac{{a}^{\mathrm{2}} }{{b}+{c}}\:+\:\frac{{b}^{\mathrm{2}} }{{c}+{a}}\:+\:\frac{{c}^{\mathrm{2}} }{{a}+{b}}\:\geqslant\:\frac{{a}+{b}+{c}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\checkmark}\\{\:\mathrm{the}\:\mathrm{inequality}\:\mathrm{is}\:\mathrm{proven}\:\:\mathrm{True}}\\\hline\end{array} \\ $$$$ \\ $$$$ \\ $$
Answered by A5T last updated on 29/Apr/25
(a^2 /(b+c))+(b^2 /(c+a))+(c^2 /(a+b))≥(((a+b+c)^2 )/(a+b+c+a+a+b))=((a+b+c)/2)
$$\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{b}+\mathrm{c}}+\frac{\mathrm{b}^{\mathrm{2}} }{\mathrm{c}+\mathrm{a}}+\frac{\mathrm{c}^{\mathrm{2}} }{\mathrm{a}+\mathrm{b}}\geqslant\frac{\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)^{\mathrm{2}} }{\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{a}+\mathrm{a}+\mathrm{b}}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{\mathrm{2}} \\ $$

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