Question

cho 3 số thực a,b,c thõa mãn abc=1.
Tìm giá trị nhỏ nhất của P=\(\dfrac{(1+a)^2+b^2+5}{ab+a+4} + \dfrac{(1+b)^2+c^2+5}{bc+b+4} +\dfrac{(1+c)^2+c^2+5}{ac+c+4} \)

Answer 1

\(\frac{\left(1+a\right)^2+b^2+5}{ab+a+4}=\frac{a^2+b^2+2a+6}{ab+a+4}\ge\frac{2ab+2a+6}{ab+a+4}=2-\frac{2}{ab+a+1+3}\ge2-\frac{1}{2}\left(\frac{1}{ab+a+1}+\frac{1}{3}\right)\)

Tương tự: \(\frac{\left(1+b\right)^2+c^2+5}{bc+b+4}\ge2-\frac{1}{2}\left(\frac{1}{bc+b+1}+\frac{1}{3}\right)\) ; \(\frac{\left(1+c\right)^2+c^2+5}{ac+c+4}\ge2-\frac{1}{2}\left(\frac{1}{ac+c+1}+\frac{1}{3}\right)\)

Cộng vế với vế:

\(P\ge\frac{11}{2}-\frac{1}{2}\left(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}\right)=\frac{11}{2}-\frac{1}{2}=5\)

\(P_{min}=5\) khi \(a=b=c=1\)