Question

If a,b and c are three positive real numbers, which one of the following are true?

• A. $a^2+b^2+c^2\ge bc+ca+ab$
• B. $a^3+b^3+c^3\ge 3abc$
• C. $(b+c)(c+a)(a+b)\ge 8abc$
• D. $\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\ge a+b+c$

Answer 1

Answer: (A), (B), (C), (D)

$\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\ge a+b+c$

We have
$\frac{a^2+b^2}{2}\ge \sqrt{a^2{b}^2}=ab,\frac{b^2+c^2}{2}\ge \sqrt{b^2{c}^2}=bcand\frac{c^2+a^2}{2}\ge \sqrt{c^2{a}^2}=ca$
Adding these inequalities, we get
$a^2+b^2+c^2\ge bc+ca+abAlso\frac{a^3+b^3+c^3}{3}\ge (a^3{b}^3{c}^3{)}^{1/3}\Rightarrow a^3+b^3+c^3\ge 3abc$
$Next,since(b+c)/2\ge \sqrt{bc},(c+a)/2\ge \sqrt{ca}and(a+b)/2\ge \sqrt{ab},weget\left(\frac{b+c}{2}\right)\left(\frac{c+a}{2}\right)\left(\frac{a+b}{2}\right)\ge \sqrt{a^2{b}^2{c}^2}\Rightarrow (b+c)(c+a)(a+b)\ge 8abcLastly,wehave\frac{1}{2}(\frac{bc}{a}+\frac{ca}{b})\ge \sqrt{\frac{bc}{a}.\frac{ca}{b}}=c,\frac{1}{2}(\frac{ca}{b}+\frac{ab}{c})\ge \sqrt{\frac{ca}{b}.\frac{ab}{c}}=b,and\frac{1}{2}(\frac{ab}{c}+\frac{bc}{a})\ge \sqrt{\frac{ab}{c}.\frac{bc}{a}}=b$
Adding the above inequalities we get
$\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\ge a+b+c$