Question

[Đặng Thiên Long]

Giải tiếp mình với !!!!!

1. Cho x,y là hai số dương thỏa x+y=1. C/M \(\frac{x}{\sqrt{1-x^2}}+\frac{y}{\sqrt{1-y^2}}\ge\frac{2}{\sqrt{3}}\)

2. Chứng minh rằng nếu \(0< b< a\le2\)và \(2ba\le2b+a\)thì \(a^2+b^2\le5\)

3. Cho ba số a,b,c dương thỏa abc=1. Chứng minh: 

\(\left(a+\frac{1}{b}-1\right)\left(b+\frac{1}{c}-1\right)\left(c+\frac{1}{a}-1\right)\le1\)

4. Cho a,b,c,d dương thỏa: \(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}+\frac{1}{1+d}=3\)

Chứng minh: \(abcd\le81\)

Answer 1

[Nguyễn Tất Đạt]

Câu 1: Đặt   \(S=\frac{x}{\sqrt{1-x^2}}+\frac{y}{\sqrt{1-y^2}}=\frac{x}{\sqrt{\left(1-x\right)\left(x+1\right)}}+\frac{y}{\sqrt{\left(1-y\right)\left(y+1\right)}}\)

\(\frac{S}{\sqrt{3}}=\frac{x}{\sqrt{\left(3-3x\right)\left(x+1\right)}}+\frac{y}{\sqrt{\left(3-3y\right)\left(y+1\right)}}\)

Áp dụng BĐT AM-GM: \(\sqrt{\left(3-3x\right)\left(x+1\right)}\le\frac{3-3x+x+1}{2}=\frac{4-2x}{2}=2-x\)

\(\Rightarrow\frac{x}{\sqrt{\left(3-3x\right)\left(x+1\right)}}\ge\frac{x}{2-x}\)

Tương tự: \(\frac{y}{\sqrt{\left(3-3y\right)\left(y+1\right)}}\ge\frac{y}{2-y}\)

Từ đó: \(\frac{S}{\sqrt{3}}\ge\frac{x}{2-x}+\frac{y}{2-y}=\frac{x^2}{2x-x^2}+\frac{y^2}{2y-y^2}\)

Áp dụng BĐT Schwarz: \(\frac{S}{\sqrt{3}}\ge\frac{x^2}{2x-x^2}+\frac{y^2}{2y-y^2}\ge\frac{\left(x+y\right)^2}{2\left(x+y\right)-\left(x^2+y^2\right)}=\frac{1}{2-\left(x^2+y^2\right)}\)

Áp dụng BĐT \(\frac{x^2+y^2}{2}\ge\frac{\left(x+y\right)^2}{4}\Rightarrow x^2+y^2\ge\frac{\left(x+y\right)^2}{2}=\frac{1}{2}\)

\(\Rightarrow\frac{S}{\sqrt{3}}\ge\frac{1}{2-\frac{1}{2}}=\frac{2}{3}\Leftrightarrow S\ge\frac{2\sqrt{3}}{3}=\frac{2}{\sqrt{3}}\)(ĐPCM).

Dấu bằng có <=> \(x=y=\frac{1}{2}\).

Answer 2

[Nguyễn Tất Đạt]

Câu 4: Sửa đề CMR: \(abcd\le\frac{1}{81}\)

 Ta có: \(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}+\frac{1}{1+d}=3\)

\(\Leftrightarrow\frac{1}{1+a}=\left(1-\frac{1}{1+b}\right)+\left(1-\frac{1}{1+c}\right)+\left(1-\frac{1}{1+d}\right)\)

\(\Leftrightarrow\frac{1}{1+a}=\frac{b}{1+b}+\frac{c}{1+c}+\frac{d}{1+d}\ge3\sqrt[3]{\frac{bcd}{\left(1+b\right)\left(1+c\right)\left(1+d\right)}}\)(AM-GM)

Tương tự: 

\(\frac{1}{1+b}\ge3\sqrt[3]{\frac{acd}{\left(1+a\right)\left(1+c\right)\left(1+d\right)}}\)\(;\frac{1}{1+c}\ge3\sqrt[3]{\frac{abd}{\left(1+a\right)\left(1+b\right)\left(1+d\right)}}\)

\(\frac{1}{1+d}\ge3\sqrt[3]{\frac{abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)

Nhân 4 BĐT trên theo vế thì có: 

\(\frac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)}\ge81\sqrt[3]{\frac{\left(abcd\right)^3}{\left[\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)\right]^3}}\)

\(=81.\frac{abcd}{\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)}\)

\(\Rightarrow81.abcd\le1\Leftrightarrow abcd\le\frac{1}{81}\)(ĐPCM)

Dấu "=" có <=> \(a=b=c=d=\frac{1}{3}\).