Giải thích các bước giải:
$a.\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}+\dfrac{1}{\sqrt{c}}=1$
$\rightarrow (\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}+\dfrac{1}{\sqrt{c}})^2=1$
$\rightarrow 3((\dfrac{1}{\sqrt{a}})^2+(\dfrac{1}{\sqrt{b}})^2+(\dfrac{1}{\sqrt{c}})^2)\ge(\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}+\dfrac{1}{\sqrt{c}})^2 $
$\rightarrow 3(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})\ge 1$
$\rightarrow \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge \dfrac{1}{3}$
$\rightarrow \dfrac{ab+bc+ca}{abc}\ge \dfrac{1}{3}$
$\rightarrow ab+bc+ca\ge \dfrac{abc}{3}$
$b.4x-3y=15$
$\rightarrow (4x-3y)^2=15^2$
$\rightarrow 15^2=(4x-3y)^2\le (4^2+(-3)^2)(x^2+y^2)=25(x^2+y^2)$
$\rightarrow x^2+y^2\ge 9$
