Question Number 13294 by Tinkutara last updated on 17/May/17
$$\mathrm{If}\:{a},\:{b}\:\mathrm{and}\:{c}\:\mathrm{are}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle} \\ $$$$\mathrm{and}\:{a}\:+\:{b}\:+\:{c}\:=\:\mathrm{2},\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$${a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \:+\:\mathrm{2}{abc}\:<\:\mathrm{2} \\ $$
Commented by prakash jain last updated on 18/May/17
$${S}={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{abc} \\ $$$$\mathrm{4}−{S}=\left({a}+{b}+{c}\right)^{\mathrm{2}} −{S} \\ $$$$=\mathrm{2}{ab}+\mathrm{2}{bc}+\mathrm{2}{ca}−\mathrm{2}{abc} \\ $$$$=\mathrm{2}\left({ab}+{bc}+{ca}−{abc}\right)\:\:\:\:…\left({A}\right) \\ $$$${a}<{b}+{c}\Rightarrow\mathrm{2}{a}<{a}+{b}+{c}\Rightarrow{a}<\mathrm{1}\Rightarrow\left(\mathrm{1}−{a}\right)>\mathrm{0} \\ $$$$\Rightarrow{a}<\mathrm{1}\:{also}\:{b}<\mathrm{1}\:,{c}<\mathrm{1} \\ $$$$\left(\mathrm{1}−{a}\right)\left(\mathrm{1}−{b}\right)\left(\mathrm{1}−{c}\right)>\mathrm{0} \\ $$$$\left(\mathrm{1}−{a}−{b}+{ab}\right)\left(\mathrm{1}−{c}\right)>\mathrm{0} \\ $$$$\mathrm{1}−{a}−{b}+{ab}−{c}+{ac}+{bc}−{abc}>\mathrm{0} \\ $$$$\mathrm{1}−\left({a}+{b}+{c}\right)+\left({ab}+{bc}+{ca}−{abc}\right)>\mathrm{0} \\ $$$$−\mathrm{1}+\left({ab}+{bc}+{ca}−{abc}\right)>\mathrm{0} \\ $$$$\Rightarrow\left({ab}+{bc}+{ca}−{abc}\right)>\mathrm{1}\:\:\:\:…\left({B}\right) \\ $$$${substituting}\:{B}\:{in}\:{A} \\ $$$$\mathrm{4}−{S}>\mathrm{2}\Rightarrow\mathrm{2}>{S}\:\blacksquare \\ $$
Commented by RasheedSindhi last updated on 18/May/17
$$\mathbb{X}\mathrm{cellent}!!! \\ $$
Commented by prakash jain last updated on 18/May/17
$$\mathrm{Thanks} \\ $$