Question

cho các số thực a,b,c>25/4 tìm minQ = \(\dfrac{a}{2\sqrt{b}-5}\) + \(\dfrac{b}{2\sqrt{c}-5}\) +\(\dfrac{c}{2\sqrt{a}-5}\)

Answer 1

Áp dụng BĐT AM-GM

\(\dfrac{a}{2\sqrt{b}-5}+2\sqrt{b}-5\ge2\sqrt{a}\)

\(\dfrac{b}{2\sqrt{c}-5}+2\sqrt{c}-5\ge2\sqrt{b}\)

\(\dfrac{c}{2\sqrt{a}-5}+2\sqrt{a}-5\ge2\sqrt{c}\)

\(\Rightarrow\left(\dfrac{a}{2\sqrt{b}-5}+\dfrac{b}{2\sqrt{c}-5}+\dfrac{c}{2\sqrt{a}-5}\right)+\left(2\sqrt{a}-5+2\sqrt{b}-5+2\sqrt{c}-5\right)\ge2\sqrt{a}+2\sqrt{b}+2\sqrt{c}\)

\(\Rightarrow Q=\dfrac{a}{2\sqrt{b}-5}+\dfrac{b}{2\sqrt{c}-5}+\dfrac{c}{2\sqrt{a}-5}\ge15\)

Dấu '=' xảy ra khi và chỉ khi \(a=b=c=25\left(thỏa.a;b;c>\dfrac{25}{4}\right)\)

Vậy \(Q\left(min\right)=15\) khi \(a=b=c=25\)