424)
C1: \(a+b=1\Rightarrow\left(a+b\right)^2=1\Leftrightarrow a^2+2ab+b^2=1\)
\(A=a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)=a^2-ab+b^2\)
\(A=\left(a+b\right)^2-3ab=1-3ab\)
\(1-3ab\ge1-\dfrac{3\left(a+b\right)^2}{4}\)
\(1-3ab\ge1-\dfrac{3}{4}\)
\(1-3ab\ge\dfrac{1}{4}\)
MinA = 1/4 khi a=b=1/2
C2:
Áp dụng BĐT BCS, ta có:
\(\left(a+b\right)^2\le\left(1^2+1^2\right)\left(a^2+b^2\right)\)
\(1\le2\left(a^2+b^2\right)\)
\(\dfrac{1}{2}\le a^2+b^2\)
\(\dfrac{1}{4}\le\left(a^2+b^2\right)^2\)
\(\dfrac{1}{4}\le\left(a^2+b^2\right)^2\left(a\sqrt{a}.\sqrt{a}+b\sqrt{b}.\sqrt{b}\right)^2\le\left(a^3+b^3\right)\left(a+b\right)=a^3+b^3\)
Min là 1/4 khi a=b=1/2
424:
ta có : \(a+b=1\Rightarrow a=1-b\)
\(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)
hay \(a^3+b^3=a^2-ab+b^2\)
thay \(a=1-b\) vào biểu thức \(a^2-ab+b^2\), ta được:
\(\left(1-b\right)^2-\left(1-b\right)b+b^2\)
\(=1-2b+b^2-b+b^2+b^2=3b^2-3b+1\)
\(=3\left(b+\dfrac{-3}{2.3}\right)^2+\dfrac{4.3.1-\left(-3\right)^2}{4.3}\ge\dfrac{4.3.1-\left(-3\right)^2}{4.3}=\dfrac{1}{4}=0,25\)
dấu "=" xảy ra khi b=0,5; a=0,5
vậy GTNN của \(a^3+b^3\) là 0,25 tại a=b=0,5
