Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(\text{VT}=\frac{a^3}{2b+3c}+\frac{b^3}{2c+3a}+\frac{c^3}{2a+3b}=\frac{a^4}{2ab+3ac}+\frac{b^4}{2bc+3ba}+\frac{c^4}{2ac+3bc}\)
\(\geq \frac{(a^2+b^2+c^2)^2}{2ab+3ac+2bc+3ba+2ac+3bc}=\frac{(a^2+b^2+c^2)^2}{5(ab+bc+ac)}\)
Theo hệ quả của BĐT AM-GM ta có:
\(a^2+b^2+c^2\geq ab+bc+ac\)
\(\Rightarrow \text{VT}\geq \frac{(a^2+b^2+c^2)(ab+bc+ac)}{5(ab+bc+ac)}=\frac{a^2+b^2+c^2}{5}\)
Ta có đpcm.
Dấu bằng xảy ra khi \(a=b=c\)
\(\dfrac{4}{a+b}-\dfrac{2a^2+3b^2}{2a^3+3b^3}-\dfrac{2b^2+3a^2}{2b^3+3a^3}=\dfrac{\left(a-b\right)^2.\left(12b^4+12ab^3-a^2b^2+12a^3b+12a^4\right)}{\left(a+b\right)\left(2a^3+3b^3\right)\left(2b^3+3a^3\right)}\ge0\)
PS: Còn cách dùng holder nữa mà lười quá
holder Câu hỏi của Lê Minh Đức - Toán lớp 9 - Học toán với OnlineMath
\(\dfrac{a^2b^2}{2a^2+b^2+3a^2b^2}=\dfrac{a^2b^2}{\left(a^2+b^2\right)+\left(a^2+a^2b^2\right)+2a^2b^2}\le\dfrac{a^2b^2}{2ab+2a^2b+2a^2b^2}=\dfrac{ab}{2\left(1+a+ab\right)}\)
Tương tự và cộng lại;
\(P\le\dfrac{1}{2}\left(\dfrac{ab}{1+a+ab}+\dfrac{bc}{1+b+bc}+\dfrac{ca}{1+c+ca}\right)\)
\(P\le\dfrac{1}{2}\left(\dfrac{ab}{1+a+ab}+\dfrac{abc}{a+ab+abc}+\dfrac{ab.ca}{ab+abc+ab.ca}\right)\)
\(P\le\dfrac{1}{2}\left(\dfrac{ab}{1+a+ab}+\dfrac{1}{a+ab+1}+\dfrac{a}{ab+1+a}\right)=\dfrac{1}{2}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
\(\dfrac{1}{a+b}+\dfrac{1}{a+c}+\dfrac{1}{b+c}+\dfrac{1}{b+c}\ge\dfrac{16}{2a+3b+3c}\)
\(\dfrac{1}{b+c}+\dfrac{1}{a+b}+\dfrac{1}{a+c}+\dfrac{1}{a+c}\ge\dfrac{16}{2b+3a+3c}\)
\(\dfrac{1}{a+c}+\dfrac{1}{b+c}+\dfrac{1}{a+b}+\dfrac{1}{a+b}\ge\dfrac{16}{2c+3a+3b}\)
cộng tất cả lại ta được \(4.2017\ge16.\left(\dfrac{1}{2a+3b+3c}+\dfrac{1}{2b+3a+3c}+\dfrac{1}{2c+3a+3b}\right)< =>P\le\dfrac{2017}{4}\)
dấu bằng xảy ra khi \(\left\{{}\begin{matrix}\dfrac{1}{a+b}=\dfrac{1}{b+c}=\dfrac{1}{a+c}\\\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}=2017\end{matrix}\right.< =>\left\{{}\begin{matrix}a=b=c\\\dfrac{3}{2a}=\dfrac{3}{2b}=\dfrac{3}{2c}=2017\end{matrix}\right.< =>a=b=c=\dfrac{3}{4034}}\)
\(\dfrac{ab}{a+3b+2c}=\dfrac{ab}{\left(a+c\right)+\left(b+c\right)+2b}\le\dfrac{ab}{9}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}+\dfrac{1}{2b}\right)=\dfrac{1}{9}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{a}{2}\right)\)
Tương tự:
\(\dfrac{bc}{b+3c+2a}\le\dfrac{1}{9}\left(\dfrac{bc}{a+b}+\dfrac{bc}{c+a}+\dfrac{b}{2}\right)\)
\(\dfrac{ca}{c+3a+2b}\le\dfrac{1}{9}\left(\dfrac{ca}{b+c}+\dfrac{ca}{a+b}+\dfrac{c}{2}\right)\)
Cộng vế:
\(VT\le\dfrac{1}{9}\left(\dfrac{bc+ca}{a+b}+\dfrac{ca+ab}{b+c}+\dfrac{bc+ab}{c+a}+\dfrac{a+b+c}{2}\right)=\dfrac{a+b+c}{6}\)
Dấu "=" xảy ra khi \(a=b=c\)
Giải:
Ta có: \(a-b=5\Leftrightarrow a=b+5\)
\(\dfrac{4a-b}{3a+5}+\dfrac{3b-a}{2b-5}=\dfrac{4b+20-b}{3b+15+5}+\dfrac{3b-b-5}{2b-5}\)
\(=\dfrac{3b+20}{3b+20}+\dfrac{2b-5}{2b-5}=1+1=2\)
Vậy...
ta có : a-b=5 => a=b+5 khi đó pt trên trở thành:
\(\dfrac{3a+a-b}{3a+5}+\dfrac{2b+b-a}{2b+5}=\dfrac{3a+5}{3a+5}+\dfrac{2b+5}{2b+5}=1+1=2\)
vậy ......