H
huradeli
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1,cho x,y >0.CM $\dfrac{1}{x^2}+\dfrac{1}{y^2}$\geq$\dfrac{8}{(x+y)^2}$
2,cho x,y,z>0;xy+yz+zx\leqxyz
CM:
$\dfrac{8}{x+y}+\dfrac{8}{y+z}+\dfrac{8}{z+x}$\leq$\dfrac{y+z}{x^2}+\dfrac{z+x}{y^2}+\dfrac{x+y}{z^2}+2$
3,cho a,b,c sao cho a>b>c
CM:
$\dfrac{a^2+c^2}{2}+\dfrac{1}{(a-b)^2}+\dfrac{1}{(b-c)^2}$\geq$ac+4$
4,cho a,b,c>0;a+b+c=1
CM:
$\dfrac{1}{(a+b)^2}+\dfrac{1}{(b+c)^2}+\dfrac{1}{(a+c)^2}$\geq$\dfrac{4}{(1+a)^2}
+\dfrac{4}{(1+b)^2}+\dfrac{4}{(1+c)^2}$
2,cho x,y,z>0;xy+yz+zx\leqxyz
CM:
$\dfrac{8}{x+y}+\dfrac{8}{y+z}+\dfrac{8}{z+x}$\leq$\dfrac{y+z}{x^2}+\dfrac{z+x}{y^2}+\dfrac{x+y}{z^2}+2$
3,cho a,b,c sao cho a>b>c
CM:
$\dfrac{a^2+c^2}{2}+\dfrac{1}{(a-b)^2}+\dfrac{1}{(b-c)^2}$\geq$ac+4$
4,cho a,b,c>0;a+b+c=1
CM:
$\dfrac{1}{(a+b)^2}+\dfrac{1}{(b+c)^2}+\dfrac{1}{(a+c)^2}$\geq$\dfrac{4}{(1+a)^2}
+\dfrac{4}{(1+b)^2}+\dfrac{4}{(1+c)^2}$
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