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\(3=a+b+ab\le a+b+\frac{1}{4}\left(a+b\right)^2\)
\(\Rightarrow\left(a+b\right)^2+4\left(a+b\right)-12\ge0\)
\(\Leftrightarrow\left(a+b-2\right)\left(a+b+6\right)\ge0\)
\(\Leftrightarrow a+b-2\ge0\Rightarrow a+b\ge2\)
Ta có:
BĐT\(\Leftrightarrow\frac{3a^2+3a+3b^2+3b}{\left(b+1\right)\left(a+1\right)}+\frac{ab}{a+b}\le a^2+b^2+\frac{3}{2}\)
\(\Leftrightarrow\frac{3a^2+3b^2+3a+3b}{4}+\frac{ab}{a+b}\le a^2+b^2+\frac{3}{2}\)
\(\Leftrightarrow3a+3b+\frac{4ab}{a+b}\le a^2+b^2+6\)
\(\Leftrightarrow3a+3b+\frac{4ab}{a+b}\le a^2+b^2+2\left(ab+a+b\right)\)
\(\Leftrightarrow a+b+\frac{4ab}{a+b}\le\left(a+b\right)^2\)
Ta có:
\(VT=a+b+\frac{4ab}{a+b}\le a+b+\frac{\left(a+b\right)^2}{a+b}=2\left(a+b\right)\le\left(a+b\right)\left(a+b\right)=\left(a+b\right)^2\)
Dấu "=" xảy ra khi \(a=b=1\)
Từ \(a+b+ab=3\Rightarrow a+b=3-ab\ge3-\frac{\left(a+b\right)^2}{4}\)
\(\Rightarrow\left(a+b+6\right)\left(a+b-2\right)\ge0\Rightarrow a+b\ge2\)
Biến đổi bài toán như sau:
\(P=\frac{3a}{b+1}+\frac{3b}{a+1}+\frac{ab}{a+b}-a^2-b^2\le\frac{3}{2}\)
Tức là chứng minh \(\frac{3}{2}\) là GTLN của \(P\)
\(P=\frac{3\left(a^2+b^2\right)+3\left(a+b\right)}{ab+a+b+1}+\frac{3-a-b}{a+b}-\left(a+b\right)^2++2\left(3-a-b\right)\)
\(=\frac{3}{4}\left[3\left(a+b\right)^2-6\left(3-a-b\right)+3\left(a+b\right)\right]\)
\(+\frac{3}{a+b}-1-\left(a+b\right)^2+6-2\left(a+b\right)\)
Khảo sat đồ thì trên \(a+b\ge2\) tìm tìm được \(P_{Max}=\frac{3}{2}\)
P/s:giờ mk đi ngủ, mệt r` chỗ nào khó hiểu mai hỏi :D
ta có: \(VT=\frac{a\left(a+b+ab\right)}{b+1}+\frac{b\left(a+b+ab\right)}{a+1}+\frac{ab}{a+b}\)
\(=a^2+b^2+\frac{ab}{a+b}+\frac{ab}{a+1}+\frac{ab}{b+1}\)
cần cm \(\frac{ab}{a+b}+\frac{ab}{a+1}+\frac{ab}{b+1}\le\frac{3}{2}\)
theo giả thiết \(4=\left(a+1\right)\left(b+1\right)\le\frac{1}{4}\left(a+b+2\right)^2\)
\(\Leftrightarrow a+b\ge2\)
ta có: \(\frac{ab}{a+b}=\frac{ab+a+b}{a+b}-1=\frac{3}{a+b}\le\frac{3}{2}-1\)(*)
\(\frac{ab}{a+1}+\frac{ab}{b+1}\le\frac{1}{4}\left(b+ab\right)+\frac{1}{4}\left(a+ab\right)=\frac{1}{4}\left(3+ab\right)\)(**)
giờ cần tìm max ab.để ý rằng \(ab=ab+a+b-\left(a+b\right)=3-\left(a+b\right)\le3-2=1\)
khi đó \(\frac{ab}{a+b}+\frac{ab}{a+1}+\frac{ab}{b+1}\le\frac{3}{2}-1+\frac{1}{4}\left(3+1\right)=\frac{3}{2}\)(đpcm)
dấu = xảy ra khi a=b=1
Với \(a>0,b>0,a\ne b\)
\(\frac{a-\sqrt{ab}+b}{a\sqrt{a}+b\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}-1}{a-b}\)
\(=\)\(\frac{a-\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}-\frac{\sqrt{a}-\sqrt{b}}{a-b}+\frac{1}{a-b}\)
\(=\frac{1}{\sqrt{a}+\sqrt{b}}-\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{1}{a-b}=\frac{1}{a-b}\)
mấy bài cơ bản nên cũng dễ, mk có thể giải hết cho bn vs 1 đk : bn đăng từng câu 1 thôi nhé !
bài 3 có thể lên gg tìm kỹ thuật AM-GM (cosi) ngược dấu
bài 8 c/m bđt phụ 5b3-a3/ab+3b2 </ 2b-a ( biến đổi tương đương)
những câu còn lại 1 nửa dùng bđt AM-GM , 1 nửa phân tích nhân tử ròi dựa vào điều kiện
1. BĐT ban đầu
<=> \(\left(\frac{1}{3}-\frac{b}{a+3b}\right)+\left(\frac{1}{3}-\frac{c}{b+3c}\right)+\left(\frac{1}{3}-\frac{a}{c+3a}\right)\ge\frac{1}{4}\)
<=>\(\frac{a}{a+3b}+\frac{b}{b+3c}+\frac{c}{c+3a}\ge\frac{3}{4}\)
<=> \(\frac{a^2}{a^2+3ab}+\frac{b^2}{b^2+3bc}+\frac{c^2}{c^2+3ac}\ge\frac{3}{4}\)
Áp dụng BĐT buniacoxki dang phân thức
=> BĐT cần CM
<=> \(\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3\left(ab+bc+ac\right)}\ge\frac{3}{4}\)
<=> \(a^2+b^2+c^2\ge ab+bc+ac\)luôn đúng
=> BĐT được CM
2) \(a+b+c\le ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}\)\(\Leftrightarrow\)\(\left(a+b+c\right)^2-3\left(a+b+c\right)\ge0\)
\(\Leftrightarrow\)\(\left(a+b+c\right)\left(a+b+c-3\right)\ge0\)\(\Leftrightarrow\)\(a+b+c\ge3\)
ko mất tính tổng quát giả sử \(a\ge b\ge c\)
Có: \(3\le a+b+c\le ab+bc+ca\le3a^2\)\(\Leftrightarrow\)\(3a^2\ge3\)\(\Leftrightarrow\)\(a\ge1\)
=> \(\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a}\le\frac{3}{1+2a}\le1\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=1\)
gt <=> \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Đặt: \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\)
=> Thay vào thì \(VT=\frac{\frac{1}{xy}}{\frac{1}{z}\left(1+\frac{1}{xy}\right)}+\frac{1}{\frac{yz}{\frac{1}{x}\left(1+\frac{1}{yz}\right)}}+\frac{1}{\frac{zx}{\frac{1}{y}\left(1+\frac{1}{zx}\right)}}\)
\(VT=\frac{z}{xy+1}+\frac{x}{yz+1}+\frac{y}{zx+1}=\frac{x^2}{xyz+x}+\frac{y^2}{xyz+y}+\frac{z^2}{xyz+z}\ge\frac{\left(x+y+z\right)^2}{x+y+z+3xyz}\)
Có BĐT x, y, z > 0 thì \(\left(x+y+z\right)\left(xy+yz+zx\right)\ge9xyz\)Ta thay \(xy+yz+zx=1\)vào
=> \(x+y+z\ge9xyz=>\frac{x+y+z}{3}\ge3xyz\)
=> Từ đây thì \(VT\ge\frac{\left(x+y+z\right)^2}{x+y+z+\frac{x+y+z}{3}}=\frac{3}{4}\left(x+y+z\right)\ge\frac{3}{4}.\sqrt{3\left(xy+yz+zx\right)}=\frac{3}{4}.\sqrt{3}=\frac{3\sqrt{3}}{4}\)
=> Ta có ĐPCM . "=" xảy ra <=> x=y=z <=> \(a=b=c=\sqrt{3}\)
Xét \(\frac{a+3}{3a+bc}+\frac{b+3}{3b+ca}+\frac{c+3}{3c+ab}\)
\(\Leftrightarrow\frac{2a+b+c}{\left(a+b+c\right)a+bc}+\frac{a+2b+c}{\left(a+b+c\right)b+ca}+\frac{a+b+2c}{\left(a+b+c\right)c+ab}\)
\(\Leftrightarrow\frac{2a+b+c}{a^2+ab+ca+bc}+\frac{a+2b+c}{ab+b^2+bc+ca}+\frac{a+b+2c}{ac+bc+c^2+ab}\)
\(\Leftrightarrow\frac{2a+b+c}{a\left(a+b\right)+c\left(a+b\right)}+\frac{a+2b+c}{b\left(b+a\right)+c\left(b+a\right)}+\frac{a+b+2c}{c\left(a+c\right)+b\left(a+c\right)}\)
\(\Leftrightarrow\frac{2a+b+c}{\left(a+b\right)\left(a+c\right)}+\frac{a+2b+c}{\left(b+a\right)\left(b+c\right)}+\frac{a+b+2c}{\left(a+c\right)\left(b+c\right)}\)
Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm
\(\Rightarrow\hept{\begin{cases}\left(a+b\right)\left(a+c\right)\le\left(\frac{2a+b+c}{2}\right)^2=\frac{\left(2a+b+c\right)^2}{4}\\\left(b+a\right)\left(b+c\right)\le\left(\frac{a+2b+c}{2}\right)^2=\frac{\left(a+2b+c\right)^2}{4}\\\left(a+c\right)\left(b+c\right)\le\left(\frac{a+b+2c}{2}\right)^2=\frac{\left(a+b+2c\right)^2}{4}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}\frac{2a+b+c}{\left(a+b\right)\left(a+c\right)}\ge\frac{4\left(2a+b+c\right)}{\left(2a+b+c\right)^2}=\frac{4}{2a+b+c}\\\frac{a+2b+c}{\left(b+a\right)\left(b+c\right)}\ge\frac{4\left(a+2b+c\right)}{\left(a+2b+c\right)^2}=\frac{4}{a+2b+c}\\\frac{a+b+2c}{\left(a+c\right)\left(b+c\right)}\ge\frac{4\left(a+b+2c\right)}{\left(a+b+2c\right)^2}=\frac{4}{a+b+2c}\end{cases}}\)
\(\Rightarrow VT\ge\frac{4}{2a+b+c}+\frac{4}{a+2b+c}+\frac{4}{a+b+2c}\)
Xét \(\frac{4}{2a+b+c}+\frac{4}{a+2b+c}+\frac{4}{a+b+2c}\)
Áp dụng bất đẳng thức cộng mẫu số
\(\Rightarrow\frac{4}{2a+b+c}+\frac{4}{a+2b+c}+\frac{4}{a+b+2c}\ge\frac{\left(2+2+2\right)^2}{2a+b+c+a+2b+c+a+b+2c}\)
\(=\frac{36}{4\left(a+b+c\right)}=\frac{36}{12}=3\)
Mà \(VT\ge\frac{4}{2a+b+c}+\frac{4}{a+2b+c}+\frac{4}{a+b+2c}\)
\(\Rightarrow VT\ge3\)
\(\Leftrightarrow\frac{a+3}{3a+bc}+\frac{b+3}{3b+ca}+\frac{c+3}{3c+ab}\ge3\left(đpcm\right)\)
Chúc bạn học tốt !!!