Đáp án:
K=4
Giải thích các bước giải:
$K=(\dfrac{a+b}{2c})^2+(\dfrac{b+c}{2d})^2+(\dfrac{c+d}{2a})^2+(\dfrac{d+a}{2b})^2\\
\rightarrow K=\dfrac{1}{4}((\dfrac{a+b}{c})^2+(\dfrac{b+c}{d})^2+(\dfrac{c+d}{a})^2+(\dfrac{d+a}{b})^2)\\
\rightarrow K=\dfrac{1}{16}((\dfrac{a+b}{c})^2+(\dfrac{b+c}{d})^2+(\dfrac{c+d}{a})^2+(\dfrac{d+a}{b})^2)(1+1+1+1)\\
\rightarrow K\ge \frac{1}{16}.(\dfrac{a+b}{c}+\dfrac{b+c}{d}+\dfrac{c+d}{a}+\dfrac{d+a}{b})^2\\
\rightarrow K\ge \dfrac{1}{16}.(\dfrac{a}{c}+\dfrac{b}{c}+\dfrac{b}{d}+\dfrac{c}{d}+\dfrac{c}{a}+\dfrac{d}{a}+\dfrac{d}{b}+\dfrac{a}{b})^2\\
\rightarrow K\ge \dfrac{1}{16}.(8\sqrt[8]{\dfrac{a}{c}.\dfrac{b}{c}.\dfrac{b}{d}.\dfrac{c}{d}.\dfrac{c}{a}.\dfrac{d}{a}.\dfrac{d}{b}.\dfrac{a}{b}})^2\\
\rightarrow K\ge 4\\
\text{Dấu = xảy ra khi a=b=c=d}$
