Ta có: `(\sqrt{3a}-\sqrt{5b})^2\ge0∀a,b\ge0`
`⇔(\sqrt{3a})^2-2.\sqrt{3a}.\sqrt{5b}+(\sqrt{5b})^2\ge0`
`⇔3a+5b\ge2.\sqrt{3a.5b}`
`⇔12\ge2\sqrt{15}.\sqrt{ab}`
`⇔12:2\sqrt{15}\ge\sqrt{ab}`
`⇔\frac{2\sqrt{15}}{5}\ge\sqrt{ab}`
`⇔(\frac{2\sqrt{15}}{5})^2\ge(\sqrt{ab})^2`
`⇔12/5\geab.`
Dấu bằng xảy ra khi: $\begin{cases}(\sqrt{3a}-\sqrt{5b})^2=0\\3a+5b=12\end{cases}$
$⇔\begin{cases}3a=5b\\3a+5b=12\end{cases}$
$⇔\begin{cases}a=\dfrac{5}{3}b\\3.\dfrac{5}{3}b=12-5b\end{cases}$
$⇔\begin{cases}a=\dfrac{5}{3}b\\5b+5b=12\end{cases}$
$⇔\begin{cases}a=\dfrac{5}{3}b\\10b=12\end{cases}$
$⇔\begin{cases}a=\dfrac{5}{3}.\dfrac{6}{5}\\b=\dfrac{12}{10}=\dfrac{6}{5}\end{cases}$
$⇔\begin{cases}a=\dfrac{5}{3}.\dfrac{6}{5}=2\\b=\dfrac{6}{5}\end{cases}$
Vậy $Max_C=$`12/5⇔a=2,b=6/5.`
