a) chứng minh rằng: a^4+b^4+c^4+d^4 ≥ 4abcd (a; b; c; d ∈ R) b) Biết x+y+z=1 chứng minh: 1/x +1/y +1/z ≥ 9

Giải thích các bước giải:

a.Ta có :
$a^4+b^4+c^4+d^4$

$=(a^4+b^4)+(c^4+d^4)$

$\ge 2a^2b^2+2c^2d^2$

$=2((ab)^2+(cd)^2)$

$\ge 2.2ab.cd$

$=4abcd$

Dấu = xảy ra khi $\begin{cases}a^2=b^2\\ c^2=d^2\\ab=cd\end{cases}$

$\to |a|=|b|=|c|=|d|$

b.Ta có :
$A=\dfrac1x+\dfrac1y+\dfrac1z$

$\to A+9=\dfrac1x+\dfrac1y+\dfrac1z+9$

$\to A+9=\dfrac1x+\dfrac1y+\dfrac1z+9(x+y+z)$

$\to A+9=(9x+\dfrac1x)+(9y+\dfrac1y)+(9z+\dfrac1z)$

$\to A+9\ge 2\sqrt{9x.\dfrac1x}+2\sqrt{9y.\dfrac1y}+2\sqrt{9z.\dfrac1z}$

$\to A+9\ge 18$

$\to A\ge 9$

$\to \dfrac1x+\dfrac1y+\dfrac1z\ge 9$