Cho các số không âm $a,b$ thỏa $a+b \leq 1$ .Chứng minh rằng :
$M= a+b+ \frac{1}{a}+\frac{1}{b}\geq  5$
Ta có $M= a+b+\frac{1}{a}+\frac{1}{b} =-3(a+b) +(4a +\frac{1}{a}) +(4b+\frac{1}{b})$.
Áp dụng bất đẳng thức Cauchy ta có :
$4a+\frac{1}{a} \geq 4 $ Dấu bằng xảy ra khi và chỉ khi $a=\frac{1}{2}$ 
Tương tự cho $b$ :ta có
$ 4b + \frac{1}{b} \geq 4$ .Hơn nữa cũng có $ -3(a+b) \geq -3 $ ,dấu bằng khi $a=b=\frac{1}{2}$.
Vậy $M \geq 5$. Dấu bằng xảy ra khi và chỉ khi $a=b=\frac{1}{2}$.
$M=(a+\frac{1}{4a})+(b+\frac{1}{4b})+\frac{3}{4}(\frac{1}{a}+\frac{1}{b})$
$\geq 1+1+\frac{3}{4}.\frac{4}{a+b}\geq 1+1+\frac{3}{4}.4=5$
Cách khác: Ta có: $\frac{1}{a}+\frac{1}{b}\ge \frac{4}{a+b}$.
Đặt $t=x+y\implies 0\leq t\leq 1$.
Khi đó: $M\ge t+\frac{4}{t}=(t+\frac{1}{t})+\frac{3}{t}\ge 2+3=5$.
Dấu $=$ xảy ra tại $a=b=\frac{1}{2}$

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