Question

Cho x,y,z,t là bốn số dương nhỏ hơn 1 thỏa mãn điều kiện: xyzt=(1−x)(1−y)(1−z)(1−t). CMR: \(x^2+y^2+z^2+t^2\ge1\)

Answer 1

Ta có:\(xyzt=\left(1-x\right)\left(1-y\right)\left(1-z\right)\left(1-t\right)\Rightarrow\dfrac{1-x}{x}.\dfrac{1-y}{y}.\dfrac{1-z}{z}\dfrac{1-t}{t}=1\)

Đặt \(\left(\dfrac{1-x}{x},\dfrac{1-y}{y},\dfrac{1-z}{z},\dfrac{1-t}{t}\right)\rightarrow\left(a,b,c,d\right)\) \(\Rightarrow abcd=1\)

Cần chứng minh \(\dfrac{1}{\left(a+1\right)^2}+\dfrac{1}{\left(b+1\right)^2}+\dfrac{1}{\left(c+1\right)^2}+\dfrac{1}{\left(d+1\right)^2}\ge1\)

Bổ đề: \(\dfrac{1}{\left(m+1\right)^2}+\dfrac{1}{\left(n+1\right)^2}\ge\dfrac{1}{mn+1}\) (*)

#cm: Áp dụng BĐT Bunyakovsky:

\(\left(mn+1\right)\left(\dfrac{m}{n}+1\right)\ge\left(m+1\right)^2\Rightarrow\dfrac{1}{\left(m+1\right)^2}\ge\dfrac{\dfrac{n}{m+n}}{mn+1}\)

Tương tự: \(\dfrac{1}{\left(n+1\right)^2}\ge\dfrac{\dfrac{m}{m+n}}{mn+1}\). Cộng theo vế ta có đpcm.

Quay lại bài toán: \(VT\ge\dfrac{1}{ab+1}+\dfrac{1}{cd+1}=\dfrac{1}{\dfrac{1}{cd}+1}+\dfrac{1}{cd+1}=1\)

Vậy ta có đpcm. Dấu =xảy ra khi a=b=c=d=1 hay \(x=y=z=t=\dfrac{1}{2}\)