dễ thôi
ta có:
\(\frac{a}{1+b^2c}=a-\frac{ab^2c}{1+b^2c};\frac{b}{1+c^2d}=b-\frac{bc^2d}{1+c^2d};\frac{c}{1+d^2a}=c-\frac{cd^2a}{1+d^2a};\frac{d}{1+a^2b}=d-\frac{da^2b}{1+a^2b}\)
áp dụng cauchy ta có:
\(b^2c+1\ge2b\sqrt{c};c^2d+1\ge2c\sqrt{d};d^2a+1\ge2d\sqrt{a};a^2b+1\ge2a\sqrt{b}\)
\(=4-\frac{ab\sqrt{c}+bc\sqrt{d}+cd\sqrt{a}+da\sqrt{b}}{2}\)
theo ông cauchy thì
\(ab\sqrt{c}\le\frac{ab\left(c+1\right)}{2};bc\sqrt{d}\le\frac{bc\left(d+1\right)}{2};cd\sqrt{a}\le\frac{cd\left(a+1\right)}{2};da\sqrt{b}\le\frac{da\left(b+1\right)}{2}\)
\(\Rightarrow4-\frac{ab\sqrt{c}+bc\sqrt{d}+cd\sqrt{a}+da\sqrt{b}}{2}\ge4-\frac{\left(abc+bcd+cda+dab\right)+\left(ab+bc+cd+da\right)}{4}\)
vẫn là ông cauchy nói là \(abc+bcd+cda+dab\le\frac{1}{16}\left(a+b+c+d\right)^3=4\)
\(ab+bc+cd+da=\left(b+d\right)\left(a+c\right)\le\frac{\left(a+b+c+d\right)^2}{4}=4\)
\(\Rightarrow4-\frac{\left(abc+bcd+cda+dab\right)+\left(ab+bc+cd+da\right)}{4}\ge4-\frac{4+4}{4}=2\)
\(\Rightarrow\frac{a}{1+b^2c}+\frac{b}{1+c^2d}+\frac{c}{1+d^2a}+\frac{d}{1+a^2b}\ge2\left(Q.E.D\right)\)
dấu bằng xảy ra khi a=b=c=d=1
\(\Rightarrow\frac{a}{1+b^2c}+\frac{b}{1+c^2d}+\frac{c}{1+d^2a}+\frac{d}{1+a^2b}\ge\left(a+b+c+d\right)-\frac{ab^2c}{2b\sqrt{c}}-\frac{bc^2d}{2c\sqrt{d}}-\frac{cd^2a}{2d\sqrt{a}}-\frac{da^2b}{2a\sqrt{b}}\)
Kiệt đừng ghi dòng cuối nhé,ko bít nó ở mô ra
mk thực ra ko ko hiểu đoạn abc +bcd + cda + dab thôi còn đoạn kia mk cx làm đc
