• \(f(1) = \frac{1}{1+1} + \frac{1}{(1+1)^2} = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}\).
• Đạo hàm: \(f'(x) = -\frac{1}{(x+1)^2} - \frac{2}{(x+1)^3}\).
• \(f'(1) = -\frac{1}{4} - \frac{2}{8} = -\frac{1}{2}\).

• \(\frac{a+2}{(a+1)^2} \geq -\frac{1}{2}a + \frac{5}{4}\)
• \(\frac{b+2}{(b+1)^2} \geq -\frac{1}{2}b + \frac{5}{4}\)
• \(\frac{c+2}{(c+1)^2} \geq -\frac{1}{2}c + \frac{5}{4}\)

mik vừa tìm dc lời giải chúc bạn học tốt !

\(\frac{\left(a+2\right)}{\left(a+1\right)^2}+\frac{\left(b+2\right)}{\left(b+1\right)^2}+\frac{\left(c+2\right)}{\left(c+1\right)^2}\)

=\(\frac{\left(\left(a+1\right)+1\right)}{\left(a+1\right)^2}+\frac{\left(\left(b+1\right)+1\right)}{\left(b+1\right)^2}+\frac{\left(\left(c+1\right)+1\right)}{\left(c+1\right)^2}\)

=\(\frac{1}{a+1}+\frac{1}{\left(a+1\right)^2}+\frac{1}{b+1}+\frac{1}{\left(b+1\right)^2}+\frac{1}{c+1}+\frac{1}{\left(c+1\right)^2}\)

=\(\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)+\left(\frac{1}{\left(a+1\right)^2}+\frac{1}{\left(b+1\right)^2}+\frac{1}{\left(c+1\right)^2}\right)\) (1)

ta có hai bđt là: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\) và \(x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}\)

CM: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\)

nhân x+y+z vào cả vế trái ta có:

\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

=\(3+\left(\frac{x}{y}+\frac{y}{x}\right)+\left(\frac{y}{z}+\frac{z}{y}\right)+\left(\frac{x}{z}+\frac{z}{x}\right)\) (2)

ta lại CM tiếp : với mọi x;y là các số thực dương

=> \(\frac{x}{y}+\frac{y}{x}\ge2\)

=\(\frac{\left(x^2+y^2\right)}{xy}\ge2\)

=\(x^2+y^2\ge2xy\)

=> \(x^2-2xy+y^2\ge0\) hay \(\left(x-y\right)^2\ge0\)

vậy \(\frac{x}{y}+\frac{y}{x}\ge2\) với mọi x;y là các số thực dương(3)

áp dụng bđt(3) vào biểu thức (2) ta có:

\(3+\left(\frac{x}{y}+\frac{y}{x}\right)+\left(\frac{y}{z}+\frac{z}{y}\right)+\left(\frac{x}{z}+\frac{z}{x}\right)\ge3+2+2+2=9\)

vậy \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\) là đúng

CM: \(x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}\)

=\(3\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)

\(3\left(x^2+y^2+z^2\right)-\left(x+y+z\right)^2\ge0\)

=\(2x^2+2y^2+2z^2-2xy-2yz-2xz\ge0\)

=\(\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(x^2-2xz+z^2\right)\ge0\)

\(\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\ge0\) là luôn đúng với mọi x;y;z

=> \(x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}\) (4)

áp dụng bđt cộng mẫu ta có:

\(\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)\ge\frac{9}{a+b+c+3}=\frac96=\frac32\) (5)

áp dụng bđt phụ

=> \(\left(\frac{1}{\left(a+1\right)^2}+\frac{1}{\left(b+1\right)^2}+\frac{1}{\left(c+1\right)^2}\right)\ge\frac{\left.\left(\frac{1}{a+1}\right.+\frac{1}{b+1}+\frac{1}{c+1}\right)^2}{3}\)

mà \(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\ge\frac32\)

=> \(\left(\frac{1}{\left(a+1\right)^2}+\frac{1}{\left(b+1\right)^2}+\frac{1}{\left(c+1\right)^2}\right)\ge\frac{\left(\frac32\right)^2}{3}=\frac{9}{\frac43}=\frac94\cdot\frac13=\frac34\) (6)

áp dụng (5) và (6) trở lại biểu thức (1)

=>\(\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)+\left(\frac{1}{\left(a+1\right)^2}+\frac{1}{\left(b+1\right)^2}+\frac{1}{\left(c+1\right)^2}\right)\ge\frac32+\frac34=\frac94\)

vậy \(\frac{\left(a+2\right)}{\left(a+1\right)^2}+\frac{\left(b+2\right)}{\left(b+1\right)^2}+\frac{\left(c+2\right)}{\left(c+1\right)^2}\ge\frac94\)