Cho a,b,c,>0 sao cho $\frac{1}{a}$+$\frac{1}{b}$+ $\frac{1}{c}$=4 CMR : P= $\frac{1}{2a+b+c}$+$\frac{1}{2b+a+c}$+ $\frac{1}{2c+a+b}$$\leq$ 2

Question

Cho a,b,c,>0 sao cho $\frac{1}{a}$+$\frac{1}{b}$+ $\frac{1}{c}$=4
CMR : P= $\frac{1}{2a+b+c}$+$\frac{1}{2b+a+c}$+ $\frac{1}{2c+a+b}$$\leq$ 2

in progress 0
Madelyn 21 phút 2021-09-27T19:30:18+00:00 1 Answers 0 views 0

Answers ( )

    0
    2021-09-27T19:32:03+00:00

    Đáp án:

    áp dụng bđt $\frac{1}{a}+\frac{1}{b}≥\frac{4}{a+b}$

    dấu = xảy ra khi a=b

    Giải thích các bước giải:

    ta có

    $\frac{1}{2a+b+c}=\frac{1}{(a+b)+(a+c)} ≤\frac{1}{4}(\frac{1}{a+b}+\frac{1}{b+c}) ≤ \frac{1}{4}(\frac{1}{4}(\frac{1}{a}+\frac{1}{b})+\frac{1}{4}(\frac{1}{a}+\frac{1}{c}))=\frac{1}{16}(\frac{2}{a}+\frac{1}{b}+\frac{1}{c})$

    tương tự ta có:

    $\frac{1}{a+2b+c} ≤ \frac{1}{16}(\frac{1}{a}+\frac{2}{b}+\frac{1}{c})$

    $\frac{1}{a+b+2c} ≤ \frac{1}{16}(\frac{1}{a}+\frac{1}{b}+\frac{2}{c})$

    suy ra $P ≤\frac{1}{16}4(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=1$

    dấu = xảy ra khi $a=b=c=\frac{3}{4}$

Leave an answer

Browse

35:5x4+1-9:3 = ? ( )