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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
\(\Rightarrow\left\{\begin{matrix}\left(a+b\right)\left(a+c\right)\le\left(\frac{2a+b+c}{2}\right)^2=\frac{\left(2a+b+c\right)^2}{4}\\\left(b+a\right)\left(b+c\right)\le\left(\frac{a+2b+c}{2}\right)^2=\frac{\left(a+2b+c\right)^2}{4}\\\left(a+c\right)\left(b+c\right)\le\left(\frac{a+b+2c}{2}\right)^2=\frac{\left(a+b+2c\right)^2}{4}\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}\frac{2a+b+c}{\left(a+b\right)\left(a+c\right)}\ge\frac{4\left(2a+b+c\right)}{\left(2a+b+c\right)^2}=\frac{4}{2a+b+c}\\\frac{a+2b+c}{\left(b+a\right)\left(b+c\right)}\ge\frac{4\left(a+2b+c\right)}{\left(a+2b+c\right)^2}=\frac{4}{a+2b+c}\\\frac{a+b+2c}{\left(a+c\right)\left(b+c\right)}\ge\frac{4\left(a+b+2c\right)}{\left(a+b+2c\right)^2}=\frac{4}{a+b+2c}\end{matrix}\right.\)
\(\Rightarrow VT\ge\frac{4}{2a+b+c}+\frac{4}{a+2b+c}+\frac{4}{a+b+2c}\)
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 \(VT\ge\frac{4}{2a+b+c}+\frac{4}{a+2b+c}+\frac{4}{a+b+2c}\)
\(\Rightarrow VT\ge3\)
\(\Leftrightarrow\frac{a+3}{3a+bc}+\frac{b+3}{3b+ca}+\frac{c+3}{3c+ab}\ge3\) ( đpcm )
Ta có:
\(3a+bc=(a+b+c)a+bc=(a+c)(a+b)\)
\(\Rightarrow \sum \frac{a+3}{3a+bc}\)\(= \sum \frac{(a+c)+(a+b)}{(a+c)(a+b)}=2 \sum \frac{1}{a+b}\geq 2.\frac{9}{2(a+b+c)}=3\)
Ta có:
\(\frac{a^3b}{a^3+b^3}-\frac{ab^3}{a^3+b^3}=\frac{ab\left(a^2-b^2\right)}{a^3+b^3}=\frac{ab\left(a-b\right)}{a^2-ab+b^2}=\frac{a-b}{\frac{a}{b}+\frac{b}{a}-1}\ge\frac{a-b}{\frac{a}{b}+\frac{a}{a}-1}=\frac{b\left(a-b\right)}{a}\)
\(\frac{b^3c}{b^3+c^3}-\frac{bc^3}{b^3+c^3}=\frac{bc\left(b^2-c^2\right)}{b^3+c^3}=\frac{bc\left(b-c\right)}{b^2-bc+c^2}=\frac{b-c}{\frac{b}{c}+\frac{c}{b}-1}\ge\frac{b-c}{\frac{a}{c}+\frac{b}{b}-1}=\frac{c\left(b-c\right)}{a}\)
\(\frac{c^3a}{c^3+a^3}-\frac{ca^3}{c^3+a^3}=\frac{ca\left(c^2-a^2\right)}{c^3+a^3}=\frac{ca\left(c-a\right)}{c^2-ca+a^2}=\frac{c-a}{\frac{c}{a}+\frac{a}{c}-1}\ge\frac{c-a}{\frac{a}{c}+\frac{a}{a}-1}=\frac{c\left(c-a\right)}{a}\)
\(\Rightarrow\frac{a^3b}{a^3+b^3}-\frac{ab^3}{a^3+b^3}+\frac{b^3c}{b^3+c^3}-\frac{bc^3}{b^3+c^3}+\frac{c^3a}{c^3+a^3}-\frac{ca^3}{c^3+a^3}\ge\frac{b\left(a-b\right)+c\left(c-a\right)+c\left(b-c\right)}{a}=\frac{ab-b^2-ac+bc}{a}=\frac{\left(a-b\right)\left(b-c\right)}{a}\ge0\)
\(\Leftrightarrow\frac{a^3b}{a^3+b^3}+\frac{b^3c}{b^3+c^3}+\frac{c^3a}{c^3+a^3}\ge\frac{ab^3}{a^3+b^3}+\frac{bc^3}{b^3+c^3}+\frac{ca^3}{c^3+a^3}\left(đpcm\right)\)
\(\frac{a+3}{3a+bc}=\frac{a+a+b+c}{\left(a+b+c\right)a+bc}=\frac{\left(a+b\right)+\left(a+c\right)}{\left(a+b\right)\left(a+c\right)}=\frac{1}{a+b}+\frac{1}{a+c}\)
Áp dụng bất đẳng thức Côsi dạng \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) là ổn.
Ta có:
\(\frac{2a^5+3b^5}{ab}\ge5a^3+10b^3-10ab^2\)
\(\Leftrightarrow\left(a-b\right)^4\left(2a+3b\right)\ge0\).Tương tự với 2 cái còn lại được:
\(\frac{2a^5+3b^5}{ab}+\frac{2b^5+3c^5}{cb}+\frac{2c^5+3a^5}{ab}\ge15\left(a^3+b^3+c^3\right)-10\left(ab^2+bc^2+ca^2\right)\)
=>Đpcm (vì ab2+bc2+ca2=3)
Dấu = khi a=b=c=1
bài 2 thì bạn áp dụng bdt cô si với lựa chọn điểm rơi hoặc bdt holder ( nó giống kiểu bunhia ngược ) . bai 1 thi ap dung cai nay \(\frac{1}{x}+\frac{1}{y}>=\frac{1}{x+y}\) câu 1 khó hơn nhưng bạn biết lựa chọn điểm rơi với áp dụng bdt phụ kia là ok .
Bài 1:Đặt VT=A
Dùng BĐT \(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\Rightarrow\frac{1}{x+y+z}\le\frac{1}{9}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)x,y,z>0\)
Áp dụng vào bài toán trên với x=a+c;y=b+a;z=2b ta có:
\(\frac{ab}{a+3b+2c}=\frac{ab}{\left(a+c\right)+\left(b+c\right)+2b}\le\frac{ab}{9}\left(\frac{1}{a+c}+\frac{1}{b+c}+\frac{1}{2b}\right)\)
Tương tự với 2 cái còn lại
\(A\le\frac{1}{9}\left(\frac{bc+ac}{a+b}+\frac{bc+ab}{a+c}+\frac{ab+ac}{b+c}\right)+\frac{1}{18}\left(a+b+c\right)\)
\(\Rightarrow A\le\frac{1}{9}\left(a+b+c\right)+\frac{1}{18}\left(a+b+c\right)=\frac{a+b+c}{6}\)
Đẳng thức xảy ra khi a=b=c
Bài 2:
Biến đổi BPT \(4\left(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{b^3}{\left(1+c\right)\left(1+a\right)}+\frac{c^3}{\left(1+a\right)\left(1+b\right)}\right)\ge3\)
\(\Rightarrow\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{b^3}{\left(1+c\right)\left(1+a\right)}+\frac{c^3}{\left(1+a\right)\left(1+b\right)}\ge\frac{3}{4}\)
Dự đoán điểm rơi xảy ra khi a=b=c=1
\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{1+b}{8}+\frac{1+c}{8}\ge\frac{3a}{4}\)
Tương tự suy ra
\(VT\ge\frac{2\left(a+b+c\right)-3}{4}\ge\frac{2\cdot3\sqrt{abc}-3}{4}=\frac{3}{4}\)
Sử dụng giả thiết a + b + c = 3, ta được: \(\frac{a^3}{3a-ab-ca+2bc}=\frac{a^3}{\left(a+b+c\right)a-ab-ca+2bc}\)\(=\frac{a^3}{a^2+2bc}\)
Tương tự ta có \(\frac{b^3}{3b-bc-ab+2ca}=\frac{b^3}{b^2+2ca}\); \(\frac{c^3}{3c-ca-bc+2ab}=\frac{c^3}{c^2+2ab}\)
Khi đó thì \(P=\frac{a^3}{a^2+2bc}+\frac{b^3}{b^2+2ca}+\frac{c^3}{c^2+2ab}+3abc\)\(=\left(a+b+c\right)-\frac{2abc}{a^2+2bc}-\frac{2abc}{b^2+2ca}-\frac{2abc}{c^2+2ab}+3abc\)\(=3+abc\left[3-2\left(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ca}+\frac{1}{c^2+2ab}\right)\right]\)\(\le3+abc\left[3-2.\frac{9}{a^2+b^2+c^2+2\left(ab+bc+ca\right)}\right]\)(Theo BĐT Bunyakovsky dạng phân thức)\(=3+abc\left[3-2.\frac{9}{\left(a+b+c\right)^2}\right]\le3+\left(\frac{a+b+c}{3}\right)^3=4\)
Đẳng thức xảy ra khi a = b = c = 1
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 !!!