Question Number 131465 by bramlexs22 last updated on 05/Feb/21
$${Let}\:{a},{b}\:{and}\:{c}\:{be}\:{three}\:{positive}\:{real}\:{numbers} \\ $$$$.\:{Prove}\:{that}\:{a}+{b}+{c}\:\leqslant\:\frac{{a}^{\mathrm{2}} +{bc}}{{b}+{c}}+\frac{{b}^{\mathrm{2}} +{ca}}{{c}+{a}}+\frac{{c}^{\mathrm{2}} +{ab}}{{a}+{b}} \\ $$
Answered by EDWIN88 last updated on 05/Feb/21
$${According}\:{to}\:{the}\:{identity}\:\frac{{a}^{\mathrm{2}} +{bc}}{{b}+{c}}−{a}=\frac{\left({a}−{b}\right)\left({a}−{c}\right)}{{b}+{c}} \\ $$$${we}\:{can}\:{change}\:{our}\:{inequality}\:{into}\:{the}\:{form} \\ $$$${x}\left({a}−{b}\right)\left({a}−{c}\right)+{y}\left({b}−{a}\right)\left({b}−{c}\right)+{z}\left({c}−{a}\right)\left({c}−{b}\right)\geqslant\mathrm{0} \\ $$$${in}\:{which}\:{x}=\frac{\mathrm{1}}{{b}+{c}}\:,\:{y}=\frac{\mathrm{1}}{{c}+{a}}\:,{z}=\frac{\mathrm{1}}{{a}+{b}} \\ $$$${WLOG}\:,{assume}\:{a}\geqslant{b}\geqslant{c}\:{then}\:{x}\leqslant{y}\leqslant{z} \\ $$$$ \\ $$