Question

Va rog! Dau coroana!
Ex 46 ​

Answer 1

Folosim inegalitatea mediilor:

$\it m_g\leq m_a \Rightarrow m_g^2\leq m_a^2\\ \\ \\ ( \sqrt{ab})^2\leq\Big(\dfrac{a+b}{2}\Big)^2 \Rightarrow ab\leq\dfrac{(a+b)^2}{4}|_{:(a+b)} \Rightarrow \dfrac{ab}{a+b}\leq\dfrac{a+b}{4}\ \ \ \ \ (1)\\ \\ \\ ( \sqrt{bc})^2\leq\Big(\dfrac{b+c}{2}\Big)^2 \Rightarrow bc\leq\dfrac{(b+c)^2}{4}|_{:(b+c)} \Rightarrow \dfrac{bc}{b+c}\leq\dfrac{b+c}{4}\ \ \ \ \ (2)$

$\it ( \sqrt{ca})^2\leq\Big(\dfrac{c+a}{2}\Big)^2 \Rightarrow ca\leq\dfrac{(c+a)^2}{4}|_{:(c+a)} \Rightarrow \dfrac{ca}{c+a}\leq\dfrac{c+a}{4}\ \ \ \ \ (3)\\ \\ \\ (1),\ (2),\ (3) \Rightarrow \dfrac{ab}{a+b}+\dfrac{bc}{b+c}+\dfrac{ca}{c+a}\leq\dfrac{a+b+b+c+c+a}{4} \Rightarrow \\ \\ \\ \Rightarrow \dfrac{ab}{a+b}+\dfrac{bc}{b+c}+\dfrac{ca}{c+a}\leq\dfrac{2(a+b+c)}{4}\Rightarrow \dfrac{ab}{a+b}+\dfrac{bc}{b+c}+\dfrac{ca}{c+a}\leq\dfrac{a+b+c}{2}$