Question

Demonstrați că
a+b/c + b+c/a + c+a/b >=6​

Answer 1

$\displaystyle\bf\\\boxed{\bf INEGALITATEA~MEDIILOR}~.\\\\\sqrt{\frac{a^2+b^2}{2}} \geq \frac{a+b}{2} \geq \sqrt{ab} \geq \frac{2}{\frac{1}{a}+\frac{1}{b}},~\forall~a,~b~\geq 0.\\\\---------------------------\\\\\frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b} \geq 6 \Leftrightarrow\\\\\frac{a}{c} +\frac{b}{c}+\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b} \geq 6 \Leftrightarrow\\\\\bigg(\frac{a}{c}+\frac{c}{a}\bigg)+\bigg(\frac{b}{c}+\frac{c}{b}\bigg)+\bigg(\frac{b}{a}+\frac{a}{b}\bigg) \geq 6.$

$\displaystyle\bf\\folosim~inegalitatea~mediilor,~M_a-M_g.\\\frac{a}{c} +\frac{c}{a} \geq 2\sqrt{\frac{a}{c}\cdot \frac{c}{a} } =2.\\analog~si~pentru~ceilalti~termeni,~asadar~vom~avea~:~\\\\\bigg(\frac{a}{c}+\frac{c}{a}\bigg)+\bigg(\frac{b}{c}+\frac{c}{b}\bigg)+\bigg(\frac{b}{a}+\frac{a}{b}\bigg) \geq 2+2+2=6.\\\\(q.e.d)$