Cho 4x+y=1 Chứng minh rằng: $4x^{2}$+ $y^{2}$ ≥ $\frac{1}{5}$

Question

Cho 4x+y=1
Chứng minh rằng: $4x^{2}$+ $y^{2}$ ≥ $\frac{1}{5}$

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Kennedy 1 tháng 2021-10-28T14:12:07+00:00 2 Answers 6 views 0

Answers ( )

    0
    2021-10-28T14:13:51+00:00

    Ta có $4x+y=1$

    $\to y = 1-4x$

    Ta có : $4x^2+y^2$

    $ = 4x^2+(1-4x)^2$

    $ = 20x^2-8y+1$

    $ 20.\bigg(x^2-\dfrac{2}{5}x+\dfrac{1}{20}\bigg)$

    $ = 20.\bigg(x^2-2.x.\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{100}\bigg)$

    $ = 20.\bigg(x-\dfrac{1}{5}\bigg)^2 +\dfrac{1}{5} ≥ \dfrac{1}{5}$

    Dấu “=” xảy ra $⇔x=y=\dfrac{1}{5}$

     

    0
    2021-10-28T14:13:58+00:00

    $4x+y=1⇒y=1-4x$

    $⇒4x^2+y^2$

    $=4x^2+(1-4x)^2$

    $=4x^2+1-8x+16x^2$

    $=20x^2-8x+1$

    $=20(x^2-\frac{2}{5}x+\frac{1}{20})$

    $=20(x-\frac{1}{5})^2+\frac{1}{5}≥\frac{1}{5}(đpcm)$.

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