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\(P=\dfrac{a}{4-3a}+\dfrac{b}{4-3b}+\dfrac{c}{4-3c}=\dfrac{a^2}{4a-3a^2}+\dfrac{b^2}{4b-3b^2}+\dfrac{c^2}{4c-3c^2}\)
\(\ge\dfrac{\left(a+b+c\right)^2}{4\left(a+b+c\right)-3\left(a^2+b^2+c^2\right)}\) (BĐT Cauchy-Schwarz)
\(=\dfrac{1}{4-3\left(a^2+b^2+c^2\right)}\)
Ta có: \(3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)
\(\Rightarrow4-3\left(a^2+b^2+c^2\right)\le4-\left(a+b+c\right)^2=4-1=3\)
\(\Rightarrow\dfrac{1}{4-3\left(a^2+b^2+c^2\right)}\ge\dfrac{1}{3}\)
\(\Rightarrow P_{min}=\dfrac{1}{3}\) khi \(a=b=c=\dfrac{1}{3}\)
Casch2:đặt \(\left\{{}\begin{matrix}4-3a=x\\4-3b=y\\4-3c=z\end{matrix}\right.\)\(=>\left\{{}\begin{matrix}a=\dfrac{4-x}{3}\\b=\dfrac{4-y}{3}\\c=\dfrac{4-z}{3}\end{matrix}\right.\)\(x+y+z=9\)
\(=>P=\dfrac{4-x}{3x}+\dfrac{4-y}{3y}+\dfrac{4-z}{3z}=\dfrac{4}{3x}+\dfrac{4}{3y}+\dfrac{4}{3z}-\left(\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}\right)\)
\(=\dfrac{\left(2+2+2\right)^2}{3.9}-1=\dfrac{4}{3}-1=\dfrac{1}{3}\)
dấu"=" xảy ra<=>x=y=z=3<=>a=b=c=1/3
a.
\(\sum\dfrac{ab}{a+c+b+c}\le\dfrac{1}{4}\sum\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)=\dfrac{a+b+c}{4}\)
2.
\(\dfrac{ab}{a+3b+2c}=\dfrac{ab}{a+b+2c+2b}\le\dfrac{ab}{9}\left(\dfrac{4}{a+b+2c}+\dfrac{1}{2b}\right)=4.\dfrac{ab}{a+b+2c}+\dfrac{a}{18}\)
Quay lại câu a
\(\dfrac{4}{a+b}-\dfrac{2a^2+3b^2}{2a^3+3b^3}-\dfrac{2b^2+3a^2}{2b^3+3a^3}=\dfrac{\left(a-b\right)^2.\left(12b^4+12ab^3-a^2b^2+12a^3b+12a^4\right)}{\left(a+b\right)\left(2a^3+3b^3\right)\left(2b^3+3a^3\right)}\ge0\)
PS: Còn cách dùng holder nữa mà lười quá
holder Câu hỏi của Lê Minh Đức - Toán lớp 9 - Học toán với OnlineMath
\(\dfrac{ab}{a+3b+2c}=\dfrac{ab}{\left(a+c\right)+\left(b+c\right)+2b}\le\dfrac{ab}{9}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}+\dfrac{1}{2b}\right)=\dfrac{1}{9}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{a}{2}\right)\)
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 \(a=b=c\)
Áp dụng bđt AM-GM:
\(\dfrac{a^3b}{c}+\dfrac{b^3c}{a}+\dfrac{c^3a}{b}+\dfrac{a^3c}{b}+\dfrac{b^3a}{c}+\dfrac{c^3b}{a}\ge6\sqrt[6]{\dfrac{a^8b^8c^8}{a^2b^2c^2}}=6\sqrt[6]{a^6b^6c^6}=6abc\)Dấu "=" xảy ra khi \(a=b=c\)
\(VT=\dfrac{a^3bc}{c+ab^2c}+\dfrac{ab^3c}{a+abc^2}+\dfrac{abc^3}{b+a^2bc}\)
\(=abc\left(\dfrac{a^2}{c+ab^2c}+\dfrac{b^2}{a+abc^2}+\dfrac{c^2}{b+a^2bc}\right)\)
Áp dụng bđt Cauchy-Schwarz dạng engel có:
\(VT\ge\dfrac{abc\left(a+b+c\right)^2}{a+b+c+abc\left(a+b+c\right)}\)\(=\dfrac{abc\left(a+b+c\right)}{1+abc}\)
Dấu "=" xảy ra khi \(a=b=c\)
Vậy...
Sai đề không bạn,tại a=b=c=2 thay vào không thỏa mãn nha