nguyenthehoanhsgs

Ta cm bat dang thuc sau
$\sqrt{\frac{x^{2}+1}{2}}$ + $\sqrt{x}$ <= x+1.
Bien doi tuong duong ta co ($\sqrt{x}$ - 1)$^{4}$ >= 0.(dung).
Ap dung ta co $\sum \sqrt{\frac{a^{2}+1}{2}}$+$\sum \sqrt{a}$<=a+b+c+3.
theo bdt cosi ta co a+b+c>=3.
Tu hai bdt tren ta co dpcm

ta co $\sqrt{xy}$ <= $\frac{x+y}{2}$ dua bdt can cm ve
A= $\sum \frac{2x^{2}+xy}{(y+z)^{2}}$>=$\frac{9}{4}$
that vay nhan ca tu va mau voi (2x$^{2}$+xy) sau do dung bdt svacxo.
bien doi tuong ta chi can cm
$\sum x^{4}$>=xy$^{3}$ +yz$^{3}$+zx$^{3}$
va $\sum (xy)^{2}$>=xyz(x+y+z).

ta cm minP = $\frac{1}{4}$
vi xy+yz+zx=xyz suy ra x, y, z >1 va $\sum \frac{1}{x}$=1
dat x=$\frac{b+c}{a}$+1, y=$\frac{c+a}{b}$+1, z=$\frac{a+b}{c}$+1 ta dua bdt can chung minh ve
$\sum \frac{a^{3}}{(a+b+c)(b+c)^{2}}$>=$\frac{1}{4}$
ta co $\sum \frac{a^{3}}{(b+c)^{2}}$>=$\frac{(a+b+c)^{3}}{4(a+b+c)^{2}}$=$\frac{a+b+c}{4}$ (bat dang thuc holder)
chia hai ve cho (a+b+c) ta co dpcm