Question

P=16a²+2b²+\(\frac{3}{a}+\frac{2}{b}\)

Tìm GTNN của P biết 2a+b≥2

Answer 1

mình cx mới biết

P = \(4a^2+b^2+12a^2+\frac{3}{2a}+\frac{3}{2a}+b^2+\frac{1}{b}+\frac{1}{b}\\ \ge\frac{\left(2a+b\right)^2}{2}+3\sqrt[3]{\frac{12a^2\cdot3\cdot3}{2\cdot2\cdot a\cdot a}}+3\sqrt[3]{\frac{b^2\cdot1\cdot1}{b\cdot b}}\\ =2+9+3=14\)

dấu bằng khi a = 1/2 ; b = 1

Answer 2

\(P=\left(4a^2+b^2\right)+\left(12a^2+\frac{3a}{2}+\frac{3a}{2}\right)+\left(b^2+\frac{1}{b}+\frac{1}{b}\right)\ge\frac{\left(2a+b\right)^2}{2}+3\sqrt[3]{12a^2.\frac{3a}{2}.\frac{3a}{2}}+3\sqrt[3]{b^2.\frac{1}{b}.\frac{1}{b}}\)

Answer 3

\(P=4a^2+b^2+12a^2+\frac{3}{2a}+\frac{3}{2a}+b^2+\frac{1}{b}+\frac{1}{b}\\ \ge\frac{\left(2a+b\right)^2}{2}+3\sqrt[3]{\frac{12a^2\cdot3\cdot3}{2a\cdot2a}}+3\sqrt[3]{\frac{b^2}{b\cdot b}}\)

\(\ge2+9+3=14\)

dấu = khi a=1/2 , b=1

Answer 4

P =\(4a^2+b^2+b^2+\frac{1}{b}+\frac{1}{b}+12a^2+\frac{3}{2a}+\frac{3}{2a}\\ \ge\frac{\left(2a+b\right)^2}{2}+3\sqrt[3]{\frac{b^2}{b\cdot b}}+3\sqrt[3]{\frac{12a^2\cdot3\cdot3}{2a\cdot2a}}\\ \ge2+3+9=14\)

dấu = khi a=1/2 , b = 1

Answer 5

P =\(4a^2+b^2+b^2+\frac{1}{b}+\frac{1}{b}+12a^2+\frac{3}{2a}+\frac{3}{2a}\\ \ge\frac{\left(2a+b\right)^2}{2}+3\sqrt[3]{\frac{b^2}{b\cdot b}}+3\sqrt[3]{\frac{12a^2\cdot3\cdot3}{2a\cdot2a}}\\ \ge2+3+9=14\)

dấu = khi a=1/2 , b = 1

Answer 6

P =\(4a^2+b^2+b^2+\frac{1}{b}+\frac{1}{b}+12a^2+\frac{3}{2a}+\frac{3}{2a}\\ \ge\frac{\left(2a+b\right)^2}{2}+3\sqrt[3]{\frac{b^2}{b\cdot b}}+3\sqrt[3]{\frac{12a^2\cdot3\cdot3}{2a\cdot2a}}\\ \ge2+3+9=14\)

dấu = khi a=1/2 , b = 1

Answer 7

P =\(4a^2+b^2+b^2+\frac{1}{b}+\frac{1}{b}+12a^2+\frac{3}{2a}+\frac{3}{2a}\\ \ge\frac{\left(2a+b\right)^2}{2}+3\sqrt[3]{\frac{b^2}{b\cdot b}}+3\sqrt[3]{\frac{12a^2\cdot3\cdot3}{2a\cdot2a}}\\ \ge2+3+9=14\)

dấu = khi a=1/2 , b = 1