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Question Number 13294 by Tinkutara last updated on 17/May/17

If a, b and c are the sides of a triangle  and a + b + c = 2, then prove that  a^2  + b^2  + c^2  + 2abc < 2

$$\mathrm{If}\:{a},\:{b}\:\mathrm{and}\:{c}\:\mathrm{are}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle} \\ $$ $$\mathrm{and}\:{a}\:+\:{b}\:+\:{c}\:=\:\mathrm{2},\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$ $${a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \:+\:\mathrm{2}{abc}\:<\:\mathrm{2} \\ $$

Commented byprakash jain last updated on 18/May/17

S=a^2 +b^2 +c^2 +2abc  4−S=(a+b+c)^2 −S  =2ab+2bc+2ca−2abc  =2(ab+bc+ca−abc)    ...(A)  a<b+c⇒2a<a+b+c⇒a<1⇒(1−a)>0  ⇒a<1 also b<1 ,c<1  (1−a)(1−b)(1−c)>0  (1−a−b+ab)(1−c)>0  1−a−b+ab−c+ac+bc−abc>0  1−(a+b+c)+(ab+bc+ca−abc)>0  −1+(ab+bc+ca−abc)>0  ⇒(ab+bc+ca−abc)>1    ...(B)  substituting B in A  4−S>2⇒2>S ■

$${S}={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{abc} \\ $$ $$\mathrm{4}−{S}=\left({a}+{b}+{c}\right)^{\mathrm{2}} −{S} \\ $$ $$=\mathrm{2}{ab}+\mathrm{2}{bc}+\mathrm{2}{ca}−\mathrm{2}{abc} \\ $$ $$=\mathrm{2}\left({ab}+{bc}+{ca}−{abc}\right)\:\:\:\:...\left({A}\right) \\ $$ $${a}<{b}+{c}\Rightarrow\mathrm{2}{a}<{a}+{b}+{c}\Rightarrow{a}<\mathrm{1}\Rightarrow\left(\mathrm{1}−{a}\right)>\mathrm{0} \\ $$ $$\Rightarrow{a}<\mathrm{1}\:{also}\:{b}<\mathrm{1}\:,{c}<\mathrm{1} \\ $$ $$\left(\mathrm{1}−{a}\right)\left(\mathrm{1}−{b}\right)\left(\mathrm{1}−{c}\right)>\mathrm{0} \\ $$ $$\left(\mathrm{1}−{a}−{b}+{ab}\right)\left(\mathrm{1}−{c}\right)>\mathrm{0} \\ $$ $$\mathrm{1}−{a}−{b}+{ab}−{c}+{ac}+{bc}−{abc}>\mathrm{0} \\ $$ $$\mathrm{1}−\left({a}+{b}+{c}\right)+\left({ab}+{bc}+{ca}−{abc}\right)>\mathrm{0} \\ $$ $$−\mathrm{1}+\left({ab}+{bc}+{ca}−{abc}\right)>\mathrm{0} \\ $$ $$\Rightarrow\left({ab}+{bc}+{ca}−{abc}\right)>\mathrm{1}\:\:\:\:...\left({B}\right) \\ $$ $${substituting}\:{B}\:{in}\:{A} \\ $$ $$\mathrm{4}−{S}>\mathrm{2}\Rightarrow\mathrm{2}>{S}\:\blacksquare \\ $$

Commented byRasheedSindhi last updated on 18/May/17

Xcellent!!!

$$\mathbb{X}\mathrm{cellent}!!! \\ $$

Commented byprakash jain last updated on 18/May/17

Thanks

$$\mathrm{Thanks} \\ $$

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