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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 <=> \(x=y=\frac{1}{2}\).
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 <=> \(a=b=c=d=\frac{1}{3}\).
Ta có:
\(A-B=\frac{a+b}{2}-\sqrt{ab}=\frac{a+b-2\sqrt{ab}}{2}=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{2}>0\)
Do đó: B < A và:
\(\frac{\left(a-b\right)^2}{8\left(A-B\right)}=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2\left(\sqrt{a}-\sqrt{b}\right)}{4\left(\sqrt{a}-\sqrt{b}\right)^2}=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{4}\)
Mà: \(\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{4}=\frac{a+b+2\sqrt{ab}}{4}=\frac{a+b}{4}+\frac{\sqrt{ab}}{2}=\frac{A+B}{2}\)
\(B< A\Rightarrow B< \frac{A+B}{2}< A\left(đpcm\right)\)
Câu 2/
Ta có:
\(\frac{\left(a-b\right)^2}{8\left(A-B\right)}=\frac{\left(a-b\right)^2}{8\left(\frac{a+b}{2}-\sqrt{ab}\right)}\)
= \(\frac{\left(a-b\right)^2}{4\left(\sqrt{a}-\sqrt{b}\right)^2}=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{4}\)
Ta cần chứng minh:
\(\sqrt{ab}< \frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{4}\)
\(\Leftrightarrow4\sqrt{ab}< \left(\sqrt{a}+\sqrt{b}\right)^2\)
\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2>0\)đúng
Cái còn lại làm tương tự
Bài 1:
Áp dụng BDT C-S ta có:
\(Q^2=\left(\sqrt{x-1}+\sqrt{y-2}\right)^2\)
\(\le\left(1+1\right)\left(x-1+y-2\right)\)
\(=2\cdot\left(x+y-3\right)=2\)
Bài 2:
THay vào .... ngại làm quá mà sắp rip mạng rồi
ĐKXĐ: \(a,b\ge0\)
Áp dụng bất đẳng thức AM-GM ta có:
\(a+b\ge2.\sqrt{ab}\)
Có: \(A=\frac{a+b}{2}\ge\frac{2.\sqrt{ab}}{2}=\sqrt{ab}=B\)
đpcm
Tham khảo nhé~
Đặt \(\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)=\left(x,y,z\right)\)
\(x+y+z\ge\frac{x^2+2xy}{2x+y}+\frac{y^2+2yz}{2y+z}+\frac{z^2+2zx}{2z+x}\)
\(\Leftrightarrow x+y+z\ge\frac{3xy}{2x+y}+\frac{3yz}{2y+z}+\frac{3zx}{2z+x}\)
\(\frac{3xy}{2x+y}\le\frac{3}{9}xy\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{3}\left(x+2y\right)\)
\(\Rightarrow\Sigma_{cyc}\frac{3xy}{2x+y}\le\frac{1}{3}\left[\left(x+2y\right)+\left(y+2z\right)+\left(z+2x\right)\right]=x+y+z\)
Dấu "=" xảy ra khi x=y=z
\(8\left(A-B\right)=4\left(a+b-2\sqrt{ab}\right)=4\left(\sqrt{a}-\sqrt{b}\right)^2\)
\(\frac{\left(a-b\right)^2}{4\left(\sqrt{a}-\sqrt{b}\right)^2}=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{4}\)
Điều phải chứng minh tương đương với:
\(\sqrt{ab}< \frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{4}< \frac{a+b}{2}\)
Ta có:
\(\left(\sqrt{a}-\sqrt{b}\right)^2>0\Leftrightarrow a-2\sqrt{ab}+b>0\Leftrightarrow a+2\sqrt{ab}+b>4\sqrt{ab}\)
\(\Leftrightarrow\left(\sqrt{a}+\sqrt{b}\right)^2>4\sqrt{ab}\Leftrightarrow\sqrt{ab}< \frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{4}\)(1)
\(\left(\sqrt{a}-\sqrt{b}\right)^2>0\Leftrightarrow a-2\sqrt{ab}+b>0\Leftrightarrow2\left(a+b\right)>a+b+2\sqrt{ab}\)
\(\Leftrightarrow\left(\sqrt{a}+\sqrt{b}\right)^2< 2\left(a+b\right)\)(2)
Từ (1) (2) suy ra đpcm.