Question

If a,b,c are positive real numbers, verify


\dfrac{a^2+bc}{b+c}+\dfrac{b^2+ac}{c+a} +\dfrac{c^2+ab}{a+b}\geq a+b+c

Collected in the board: Inequality

Steven Zheng Steven Zheng posted 1 month ago


Answer

\dfrac{a^2+bc}{b+c}-a+\dfrac{b^2+ac}{c+a}-b +\dfrac{c^2+ab}{a+b}-c

=\dfrac{a^2+bc-ab-ac}{b+c}+\dfrac{b^2+ac-bc-ba}{c+a}+\dfrac{c^2+ab-ca-cb}{a+b}

=\dfrac{(a-b)(a-c)}{b+c}+\dfrac{(b-c)(b-a)}{c+a}+\dfrac{(c-a)(c-b)}{a+b}

=\dfrac{(a^2-b^2)(a^2-c^2)+(b^2-c^2)(b^2-a^2)+(c^2-a^2)(c^2-b^2)}{(b+c)(c+a)(a+b)}

=\dfrac{a^4+b^4+c^4-a^2b^2-b^2c^2-c^2a^2}{(b+c)(c+a)(a+b)}

=\dfrac{(a^2-b^2)^2+(b^2-c^2)^2+(c^2-a^2)^2}{2(b+c)(c+a)(a+b)} \geq 0

Therefore,

\dfrac{a^2+bc}{b+c}+\dfrac{b^2+ac}{c+a} +\dfrac{c^2+ab}{a+b}\geq a+b+c


Steven Zheng Steven Zheng posted 1 month ago

Scroll to Top