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![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng Bất đẳng thức Cauchy Schwarz dạng Engel ta có :
\(\frac{a^2}{a+2b}+\frac{b^2}{b+2a}\ge\frac{\left(a+b\right)^2}{a+2b+b+2a}=\frac{\left(a+b\right)^2}{3\left(a+b\right)}\)
\(2\left(\frac{a^2}{2a+b}+\frac{b^2}{2b+a}\right)\ge2\left(\frac{\left(a+b\right)^2}{2a+b+2b+a}\right)=2.\frac{\left(a+b\right)^2}{3\left(a+b\right)}\)
Cộng theo vế các bất đẳng thức cùng chiều ta được :
\(\left(\frac{a^2}{a+2b}+\frac{b^2}{b+2a}\right)+2\left(\frac{a^2}{2a+b}+\frac{b^2}{2b+a}\right)\ge\frac{\left(a+b\right)^2}{3\left(a+b\right)}+2.\frac{\left(a+b\right)^2}{3\left(a+b\right)}\)
Vậy ta có ngay điều phải chứng minh
![](https://rs.olm.vn/images/avt/0.png?1311)
Mình nhầm, phải là \(\le\frac{1}{3}\)mọi người làm giúp mình với mình cần gấp
Theo BĐT Cauchy Schwarz và các biến đổi cơ bản ta dễ có được:
\(\frac{a^2}{\left(2a+b\right)\left(2a+c\right)}=\frac{a^2}{2a\left(a+b+c\right)+2a^2+bc}=\frac{1}{9}\left[\frac{\left(2a+a\right)^2}{2a\left(a+b+c\right)+2a^2+bc}\right]\)
\(\le\frac{1}{9}\left[\frac{4a^2}{2a\left(a+b+c\right)}+\frac{a^2}{2a^2+bc}\right]=\frac{1}{9}\left(\frac{2a}{a+b+c}+\frac{a^2}{2a^2+bc}\right)\)
\(\Rightarrow LHS\le\frac{1}{9}\left(2+\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}\right)\)
Tiếp tục theo BĐT Cauchy Schwarz dạng Engel:
\(\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ca}+\frac{c^2}{c^2+2ab}\ge\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2}=1\)
Ta thực hiện phép đổi biến thì:
\(\frac{ab}{ab+2c^2}+\frac{bc}{bc+2a^2}+\frac{ca}{ca+2b^2}\ge1\)
Đến đây là phần của bạn
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\frac{bc}{2ab+ac}+\frac{ca}{2ab+bc}+\frac{4ab}{bc+ca}\)
Xét \(Q=P+3=\frac{bc}{2ab+ac}+1+\frac{ca}{2ab+bc}+1+\frac{4ab}{bc+ca}+1\)
\(Q=\frac{2ab+ac+bc}{2ab+ac}+\frac{2ab+ac+bc}{2ab+bc}+\frac{4ab+bc+ca}{bc+ca}\)
\(=\left(2ab+ac+bc\right)\left(\frac{1}{2ab+ac}+\frac{1}{2ab+bc}\right)+\frac{4ab+bc+ca}{bc+ca}\)
\(\ge\left(2ab+ac+bc\right)\frac{4}{4ab+ac+bc}+\frac{4ab+bc+ca}{bc+ca}=K\)(Áp dụng BĐT \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)với a, b không âm)
\(K=\frac{2\left(4ab+ac+bc\right)+2\left(ac+bc\right)}{4ab+ac+bc}+\frac{2\left(4ab+bc+ca\right)}{9\left(ac+bc\right)}\)\(+\frac{7\left(4ab+bc+ca\right)}{9\left(ac+bc\right)}\)
\(=2+\left[\frac{2\left(ac+bc\right)}{4ab+ac+bc}+\frac{2\left(4ab+bc+ca\right)}{9\left(ac+bc\right)}\right]+\frac{7}{9}+\frac{7}{9}.\frac{4ab}{ac+bc}\)
\(\ge2+2\sqrt{\frac{2\left(ac+bc\right)}{4ab+ac+bc}.\frac{2\left(4ab+bc+ca\right)}{9\left(ac+bc\right)}}+\frac{7}{9}+\frac{7}{9}.\frac{4ab}{ac+bc}\)(Áp dụng BĐT Cô - si cho 2 số không âm)
\(=\frac{37}{9}+\frac{7}{9}.\frac{4ab}{ac+bc}\)
Mặt khác: \(6=2\left(\frac{a}{b}+\frac{b}{a}\right)+c\left(\frac{a}{b^2}+\frac{b}{a^2}\right)=\frac{2\left(a^2+b^2\right)}{ab}+\frac{c\left(a^3+b^3\right)}{a^2b^2}\)
\(=\frac{2\left(a^2+b^2\right)}{ab}+\frac{c\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2b^2}\)\(\ge\frac{2.2ab}{ab}+\frac{c\left(a+b\right)\left(2ab-ab\right)}{a^2b^2}=4+\frac{ac+bc}{ab}\)(theo BĐT \(a^2+b^2\ge2ab\))
\(\Rightarrow\frac{ac+bc}{ab}\le2\Leftrightarrow\frac{ab}{ac+bc}\ge\frac{1}{2}\)
\(\Rightarrow K\ge\frac{37}{9}+\frac{7}{9}.\frac{4ab}{ac+bc}\ge\frac{37}{9}+\frac{7}{9}.\frac{4}{2}=\frac{17}{3}\)
Ta có \(Q=P+3\ge K\ge\frac{17}{3}\Rightarrow P\ge\frac{17}{3}-3=\frac{8}{3}\)
Đẳng thức xảy ra khi \(\hept{\begin{cases}2ab+ac=2ab+bc\\\frac{2\left(ac+bc\right)}{4ab+ac+bc}=\frac{2\left(4ab+bc+ca\right)}{9\left(ac+bc\right)}\\a=b\end{cases}}\)\(\Leftrightarrow a=b=c\)
Từ \(2\left(\frac{a}{b}+\frac{b}{a}\right)+c\left(\frac{a}{b^2}+\frac{b}{a^2}\right)=6\Rightarrow6=\frac{c\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2b^2}+\frac{2\left(a^2+b^2\right)}{ab}\)
ta có \(a^2+b^2\ge2ab\Rightarrow6=\frac{c\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2b^2}+\frac{2\left(a^2+b^2\right)}{ab}\ge\frac{c\left(a+b\right)}{ab}+4\)
\(\Rightarrow0< \frac{c\left(a+b\right)}{ab}\le2\)
Lại có
\(\frac{bc}{a\left(2b+c\right)}+\frac{ac}{b\left(2a+c\right)}=\frac{\left(bc\right)^2}{abc\left(2b+c\right)}+\frac{\left(ac\right)^2}{abc\left(2a+c\right)}\ge\frac{\left(bc+ac\right)^2}{2abc\left(a+b+c\right)}\)\(=\frac{\left[c\left(a+b\right)\right]^2}{2abc\left(a+b+c\right)}\)
và \(abc\left(a+b+c\right)=ab\cdot bc+bc\cdot ba+ab\cdot ca\le\frac{\left(ab+bc+ca\right)^2}{3}\)
\(\Rightarrow\frac{bc}{a\left(2b+c\right)}+\frac{ac}{b\left(2a+c\right)}\ge\frac{3}{2}\left(\frac{c\left(a+b\right)}{ab+bc+ca}\right)^2=\frac{3}{2}\left(\frac{\frac{c\left(a+b\right)}{ab}}{1+\frac{c\left(a+b\right)}{ab}}\right)^2\)
Đặt \(t=\frac{c\left(a+b\right)}{ab}\Rightarrow P\ge\frac{3t^2}{2\left(1+t\right)^2}+\frac{4}{t}\left(0< t\le2\right)\)
Có \(\frac{3t^2}{2\left(1+t\right)^2}+\frac{4}{t}=\left(\frac{3t^2}{\left(1+t\right)^2}+\frac{4}{t}-\frac{8}{3}\right)+\frac{8}{3}=\frac{-7t^2-8t^2+32t+24}{6t\left(1+t\right)^2}+\frac{8}{3}\)
\(=\frac{\left(t-2\right)\left(-7t^2-22t-12\right)}{6t\left(1+t\right)^2}\ge0\forall t\in(0;2]\)
=> \(\frac{\left(t-2\right)\left(-7t^2-22t-12\right)}{6t\left(1+t\right)^2}+\frac{8}{3}\ge\frac{8}{3}\forall t\in(0;2]\frac{1}{2}\)
Dấu "=" xảy ra <=> t=2 hay a=b=c
![](https://rs.olm.vn/images/avt/0.png?1311)
:( Đại Ka ơi a up câu nào khó hơn đi :( :v
Solution:
Vế trái có tính thuần nhất theo 3 biến nên ta chuẩn hóa a+b+c=3.
Điểm rơi: a=b=c=1.
Khi đó:
\(A=Sigma\frac{\left(3+a\right)^2}{2a^2+\left(3-a\right)^2}\)(em ko biết kí hiệu tổng sigma ạ :v)
\(3A\Rightarrow Sigma\frac{\left(3+a\right)^2}{a^2-2a+3}\)
UCT :v
Ta cần tìm m và n sao cho
\(\frac{\left(3+a\right)^2}{a^2-2a+3}\le ma+n\) (Luôn đúng với 0<a<3)
Với điểm rơi a=1 ta có m+n=8 => n=8-m.
Ta tìm m sao cho: \(\frac{\left(3+a\right)^2}{a^2-2a+3}\le m\left(a-1\right)+8\) (luôn đúng với 0<a<3).
Đến đây giải ra ta tìm được m=4 và n=4
Ta dễ dàng cm được: \(\frac{\left(3+a\right)^2}{a^2-2a+3}\le4\left(a+1\right)\)(với o<a<3) ( cái này chứng minh tương đg) :v
Suy ra \(3A=Sigma\frac{\left(3+a\right)^2}{a^2-2a+3}\le4\left(a+b+c\right)=24\)
=> a<=8
Max A=8 <=> a=b=c=1
UCT => ez nha anh :)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{\left(1-2a\right)\left(1-2b\right)}{\left(1-a\right)\left(1-b\right)}-\frac{4\left(1-a-b\right)^2}{\left(2-a-b\right)^2}=\frac{\left(1-2a\right)\left(1-2b\right)\left(2-a-b\right)^2-4\left(1-a\right)\left(1-b\right)\left(1-a-b\right)^2}{\left(1-a\right)\left(1-b\right)\left(2-a-b\right)^2}\)
\(=\frac{2a^3-2a^2b-3a^2-2ab^2+6ab+2b^3-3b^2}{\left(1-a\right)\left(1-b\right)\left(2-a-b\right)^2}\)
\(=\frac{\left(2a^3-4a^2b+2ab^2\right)+\left(2a^2b-4ab^2+2b^3\right)-3\left(a^2-2ab+3b^2\right)}{\left(1-a\right)\left(1-b\right)\left(2-a-b\right)^2}\)
\(=\frac{2a\left(a^2-2ab+b^2\right)+2b\left(a^2-2ab+b^2\right)-3\left(a^2-2ab+b^2\right)}{\left(1-a\right)\left(1-b\right)\left(2-a-b\right)^2}\)
\(=\frac{\left(a-b\right)^2\left(2a+2b-3\right)}{\left(1-a\right)\left(1-b\right)\left(2-a-b\right)^2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng BĐT AM-GM ta có:
\(6=2\left(\frac{a}{b}+\frac{b}{a}\right)+c\left(\frac{a}{b^2}+\frac{b}{a^2}\right)\)
\(\ge4+\frac{c\left(a^3+b^3\right)}{a^2b^2}\ge4+\frac{c\left(a+b\right)}{ab}\)\(\Rightarrow\frac{c\left(a+b\right)}{ab}\in\text{(}0;2\text{]}\)
Áp dụng BĐT Cauchy-Schwarz lại có:
\(P\ge\frac{\left(bc+ca\right)^2}{2abc\left(a+b+c\right)}+\frac{4}{\frac{c\left(a+b\right)}{ab}}\)\(\ge\frac{3c^2\left(a+b\right)^2}{2\left(ab+bc+ca\right)}+\frac{4}{\frac{c\left(a+b\right)}{ab}}\)
\(=\frac{\frac{3c^2\left(a+b\right)^2}{a^2b^2}}{2\left(1+\frac{ca}{ab}+\frac{bc}{ab}\right)^2}+\frac{4}{\frac{c\left(a+b\right)}{ab}}\)
\(=\frac{\frac{3c^2\left(a+b\right)^2}{a^2b^2}}{2\left[1+\frac{c\left(a+b\right)}{ab}\right]^2}+\frac{4}{\frac{c\left(a+b\right)}{ab}}\)
Đặt \(x=\frac{c\left(a+b\right)}{ab}\left(x\in\text{(}0;2\text{]}\right)\) khi đó ta có:
\(P\ge\frac{3x^2}{2\left(1+x\right)^2}+\frac{4}{x}\) cần chứng minh \(P\ge\frac{8}{3}\Leftrightarrow\left(x-2\right)\left(7x^2+22x+12\right)\le0\forall x\in\text{(0;2]}\)
Vậy \(Min_P=\frac{8}{3}\) khi a=b=c=2