b1:choa;b;c>0.CMR\(\frac{a}{b}\)+\(\frac{b}{c}\)+\(\frac{c}{a}\)\(\ge\)\(\frac{a+b}{b+c}\)+\(\frac{b+c}{a+b}\)+1b2:cho x;y;z>1 tm \(\frac{1}{x}\)+\(\frac{1}{y}\)+\(\frac{1}{z}\)=2.CMR\(\sqrt{x+y+z}\)\(\ge\)\(\sqrt{x-1}\)+\(\sqrt{y-1}\)+\(\sqrt{z-1}\)b3:cho x;y;z>0 tm...
Đọc tiếp
b1:choa;b;c>0.CMR\(\frac{a}{b}\)+\(\frac{b}{c}\)+\(\frac{c}{a}\)\(\ge\)\(\frac{a+b}{b+c}\)+\(\frac{b+c}{a+b}\)+1
b2:cho x;y;z>1 tm \(\frac{1}{x}\)+\(\frac{1}{y}\)+\(\frac{1}{z}\)=2.CMR\(\sqrt{x+y+z}\)\(\ge\)\(\sqrt{x-1}\)+\(\sqrt{y-1}\)+\(\sqrt{z-1}\)
b3:cho x;y;z>0 tm x+y+z=xyz.CMR \(\frac{1}{\sqrt{1+x^2}}\)+\(\frac{1}{\sqrt{1+y^2}}\)+\(\frac{1}{\sqrt{1+Z^2}}\)\(\le\)\(\frac{3}{2}\)
b2 \(\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}=\sqrt{x}.\sqrt{1-\frac{1}{x}}+\sqrt{y}.\)\(\sqrt{y}.\sqrt{1-\frac{1}{y}}+\sqrt{z}.\sqrt{1-\frac{1}{z}}\)rồi dung bunhia là xong
A= \(\frac{1}{a^3}\)+ \(\frac{1}{b^3}\)+ \(\frac{1}{c^3}\)+ \(\frac{ab^2}{c^3}\)+ \(\frac{bc^2}{a^3}\)+ \(\frac{ca^2}{b^3}\)
Svacxo:
3 cái đầu >= \(\frac{9}{a^3+b^3+c^3}\)
3 cái sau >= \(\frac{\left(\sqrt{a}b+\sqrt{c}b+\sqrt{a}c\right)^2}{a^3+b^3+c^3}\)
Cô-si: cái tử bỏ bình phương >= 3\(\sqrt{abc}\)
=> cái tử >= 9abc= 9 vì abc=1
Còn lại tự làm