Easy Algebraic Inequality

Algebra Level 2

Let a , b a,b and c c be nonnegative real numbers such that a ≥ b + c a\geq b+c . Find the minimum value of a b + c + b c + a + c a + b + 2 ( a b + b c + c a ) a 2 + b 2 + c 2 . \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{2(ab+bc+ca)}{a^2+b^2+c^2}.


The answer is 3.

ChengYiin Ong by ChengYiin Ong , 17, Malaysia

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1 solution

ChengYiin Ong
May 12, 2021

We have by Cauchy Schwarz Inequality and AM-GM, a b + c + b c + a + c a + b + 2 ( a b + b c + c a ) a 2 + b 2 + c 2 ⩾ ( a + b + c ) 2 2 ( a b + b c + c a ) + 2 ( a b + b c + c a ) a 2 + b 2 + c 2 = a 2 + b 2 + c 2 2 ( a b + b c + c a ) + 2 ( a b + b c + c a ) a 2 + b 2 + c 2 + 1 ≥ 3. \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{2(ab+bc+ca)}{a^2+b^2+c^2}\geqslant \frac{(a+b+c)^2}{2(ab+bc+ca)}+\frac{2(ab+bc+ca)}{a^2+b^2+c^2}=\frac{a^2+b^2+c^2}{2(ab+bc+ca)}+\frac{2(ab+bc+ca)}{a^2+b^2+c^2}+1\ge 3.

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