x+y=1 nên \(\left(\right. 1 + \frac{1}{x} \left.\right) \left(\right. 1 + \frac{1}{y} \left.\right) - 9\)

\(= \frac{\left(\right. x + 1 \left.\right) \left(\right. y + 1 \left.\right) - 9 x y}{x y} = \frac{2 - 8 x y}{x y}\)  

\(= \frac{2 \left(\right. 1 - 4 x y \left.\right)}{x y} = \frac{2 \left(\right. \left(\right. x + y \left.\right)^{2} - 4 x y \left.\right)}{x y}\)

\(= \frac{2 \left(\right. x - y \left.\right)^{2}}{x y} \geq 0\)

Dấu đẳng thức xảy ra khi và chỉ khi \(x = y = \frac{1}{2}\).

x+y=1 nên \(\left(\right. 1 + \frac{1}{x} \left.\right) \left(\right. 1 + \frac{1}{y} \left.\right) - 9\)

\(= \frac{\left(\right. x + 1 \left.\right) \left(\right. y + 1 \left.\right) - 9 x y}{x y} = \frac{2 - 8 x y}{x y}\)  

\(= \frac{2 \left(\right. 1 - 4 x y \left.\right)}{x y} = \frac{2 \left(\right. \left(\right. x + y \left.\right)^{2} - 4 x y \left.\right)}{x y}\)

\(= \frac{2 \left(\right. x - y \left.\right)^{2}}{x y} \geq 0\)

Dấu đẳng thức xảy ra khi và chỉ khi \(x = y = \frac{1}{2}\).

x+y=1 nên \(\left(\right. 1 + \frac{1}{x} \left.\right) \left(\right. 1 + \frac{1}{y} \left.\right) - 9\)

\(= \frac{\left(\right. x + 1 \left.\right) \left(\right. y + 1 \left.\right) - 9 x y}{x y} = \frac{2 - 8 x y}{x y}\)  

\(= \frac{2 \left(\right. 1 - 4 x y \left.\right)}{x y} = \frac{2 \left(\right. \left(\right. x + y \left.\right)^{2} - 4 x y \left.\right)}{x y}\)

\(= \frac{2 \left(\right. x - y \left.\right)^{2}}{x y} \geq 0\)

Dấu đẳng thức xảy ra khi và chỉ khi \(x = y = \frac{1}{2}\).

x+y=1 nên \(\left(\right. 1 + \frac{1}{x} \left.\right) \left(\right. 1 + \frac{1}{y} \left.\right) - 9\)

\(= \frac{\left(\right. x + 1 \left.\right) \left(\right. y + 1 \left.\right) - 9 x y}{x y} = \frac{2 - 8 x y}{x y}\)  

\(= \frac{2 \left(\right. 1 - 4 x y \left.\right)}{x y} = \frac{2 \left(\right. \left(\right. x + y \left.\right)^{2} - 4 x y \left.\right)}{x y}\)

\(= \frac{2 \left(\right. x - y \left.\right)^{2}}{x y} \geq 0\)

Dấu đẳng thức xảy ra khi và chỉ khi \(x = y = \frac{1}{2}\).

x+y=1 nên \(\left(\right. 1 + \frac{1}{x} \left.\right) \left(\right. 1 + \frac{1}{y} \left.\right) - 9\)

\(= \frac{\left(\right. x + 1 \left.\right) \left(\right. y + 1 \left.\right) - 9 x y}{x y} = \frac{2 - 8 x y}{x y}\)  

\(= \frac{2 \left(\right. 1 - 4 x y \left.\right)}{x y} = \frac{2 \left(\right. \left(\right. x + y \left.\right)^{2} - 4 x y \left.\right)}{x y}\)

\(= \frac{2 \left(\right. x - y \left.\right)^{2}}{x y} \geq 0\)

Dấu đẳng thức xảy ra khi và chỉ khi \(x = y = \frac{1}{2}\).

1) \(a^{2} - a b + b^{2}\)

\(= \left(\right. a^{2} - 2 \cdot a \cdot \frac{1}{2} b + \frac{1}{4} b^{2} \left.\right) + \frac{3}{4} b^{2} = \left(\left(\right. a - \frac{1}{2} b \left.\right)\right)^{2} + \frac{3}{4} b^{2} \geq 0 \forall a , b\)

Dấu "=" xảy ra khi: \(\left{\right. a - \frac{1}{2} b = 0 \\ b = 0 \Leftrightarrow a = b = 0\) 

2) \(a^{2} - a b + b^{2} \geq \frac{1}{4} \left(\left(\right. a + b \left.\right)\right)^{2}\)

\(< = > a^{2} - a b + b^{2} \geq \frac{1}{4} \left(\right. a^{2} + 2 a b + b^{2} \left.\right) < = > a^{2} - a b + b^{2} \geq \frac{1}{4} a^{2} + \frac{1}{2} a b + \frac{1}{4} b^{2} < = > \frac{3}{4} a^{2} - \frac{3}{2} a b + \frac{3}{4} b^{2} \geq 0 < = > \frac{3}{4} \left(\right. a^{2} - 2 a b + b^{2} \left.\right) \geq 0 < = > \frac{3}{4} \left(\left(\right. a - b \left.\right)\right)^{2} \geq 0 (\text{lu} \hat{\text{o}} \text{n}\&\text{nbsp};đ \overset{ˊ}{\text{u}} \text{ng})\)

Dấu "=" xảy ra khi: \(a - b = 0 < = > a = b\)

x(x - y)(x - z) + y(y - z)(y - x) + z(z - x)(z - y) \geq 0 với điều kiện . Ta xét các trường hợp sau:

Trường hợp 1: 

Khi , ta có

:S= x(x - y)^2 \geq 0.

Trường hợp 2: 

Khi , ta có:

S= y(y - x)^2 \geq 0.

Trường hợp 3: 

Khi , ta có

S= x^2(x - y) + y^2(y - x) = (x - y)(x^2 - y^2) = -(x - y)^2(x + y) \geq 0. 

Vì cả ba trường hợp đều thoả mãn, nên bất đẳng thức luôn đúng.