Question

Urgent dau multe puncte.

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

1)

$\it m_g\leq m_a \Rightarrow m_g^2\leq m_a^2\ \ \ \ \ (*)\\ \\ \\ (*) \Rightarrow\begin{cases}\it ab\leq\dfrac{(a+b)^2}{4}\Big|_{\cdot\dfrac{1}{a+b}} \Rightarrow \dfrac{ab}{a+b}\leq\dfrac{(a+b)^2}{4(a+b)}=\dfrac{a+b}{4}\\ \\ \\ \it bc\leq\dfrac{(b+c)^2}{4}\Big|_{\cdot\dfrac{1}{b+c}} \Rightarrow \dfrac{bc}{b+c}\leq\dfrac{(b+c)^2}{4(b+c)}=\dfrac{b+c}{4}\\ \\ \\ \it ca\leq\dfrac{(c+a)^2}{4}\Big|_{\cdot\dfrac{1}{c+a}} \Rightarrow \dfrac{ca}{c+a}\leq\dfrac{(c+a)^2}{4(c+a)}=\dfrac{c+a}{4}\end{cases}$

$\it \Rightarrow \dfrac{ab}{a+b}+ \dfrac{bc}{b+c}+ \dfrac{ca}{c+a}\leq \dfrac{a+b+b+c+c+a}{4}= \dfrac{2(a+b+c)}{4}= \dfrac{a+b+c}{2}$

3)

$\it \dfrac{a^3+b^3+c^3}{3}\geq\sqrt[3]{a^3b^3c^3} \Rightarrow a^3+b^3+c^3\geq3abc$

2)

$\it \left.\begin{aligned}a+b\geq2\sqrt{ab}\\ \\ b+c\geq2\sqrt{bc}\\ \\ c+a\geq2\sqrt{ca} \end{aligned}\right\} \Rightarrow (a+b)(b+c)(c+a)\geq8\sqrt{ab\cdot bc\cdot ca}=8\sqrt{a^2b^2c^2}=8abc$