Cho $a,b,c \in \left[\frac{1}{2};2 \right].$ Chứng minh: $\left(a+b+b \right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)\leq \frac{225}{16}$
Ta có: $$a \in [\frac{1}{2};2] \Rightarrow (a- \frac{1}{2})(a-2) \leq 0$$
$$a^2 - \frac{5}{2}a + 1 \leq 0$$
$$\Rightarrow a+ \frac{1}{a} \leq \frac{5}{2}$$
Tương tự, ta có: $$\Rightarrow b+ \frac{1}{b} \leq \frac{5}{2}$$
$$\Rightarrow c+ \frac{1}{c} \leq \frac{5}{2}$$
Theo BĐT AM-GM:
$$(a+b+c)(\frac{1}{a}+ \frac{1}{b}+ \frac{1}{c}) \leq \frac{1}{4}(a+b+c+\frac{1}{a}+ \frac{1}{b}+ \frac{1}{c})^2 \leq \frac{1}{4}(3. \frac{5}{2})^2 = \frac{225}{16}$$
$$a^2 - \frac{5}{2}a + 1 \leq 0$$
$$\Rightarrow a+ \frac{1}{a} \leq \frac{5}{2}$$
Tương tự, ta có: $$\Rightarrow b+ \frac{1}{b} \leq \frac{5}{2}$$
$$\Rightarrow c+ \frac{1}{c} \leq \frac{5}{2}$$
Theo BĐT AM-GM:
$$(a+b+c)(\frac{1}{a}+ \frac{1}{b}+ \frac{1}{c}) \leq \frac{1}{4}(a+b+c+\frac{1}{a}+ \frac{1}{b}+ \frac{1}{c})^2 \leq \frac{1}{4}(3. \frac{5}{2})^2 = \frac{225}{16}$$
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