BĐT cơ bản
\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)
\(\dfrac{ab}{c+1}=ab\dfrac{1}{c+a+b+c}=ab\dfrac{1}{\left(c+a\right)+\left(b+c\right)}\le\dfrac{ab}{4}\left[\dfrac{1}{c+a}+\dfrac{1}{b+c}\right]\)
\(\dfrac{bc}{a+1}=bc\dfrac{1}{a+a+b+c}=bc\dfrac{1}{\left(a+b\right)+\left(a+c\right)}\le\dfrac{bc}{4}\left[\dfrac{1}{a+b}+\dfrac{1}{a+c}\right]\)
\(\dfrac{ac}{b+1}=ac\dfrac{1}{b+a+b+c}=ac\dfrac{1}{\left(b+a\right)+\left(b+c\right)}\le\dfrac{ac}{4}\left[\dfrac{1}{b+a}+\dfrac{1}{b+c}\right]\)
Công lại:
\(A\le\left[\dfrac{ab+bc}{4\left(c+a\right)}+\dfrac{ab+ac}{4\left(b+c\right)}+\dfrac{bc+ac}{4\left(b+a\right)}\right]\)
\(A\le\left[\dfrac{b\left(a+c\right)}{4\left(c+a\right)}+\dfrac{a\left(b+c\right)}{4\left(b+c\right)}+\dfrac{c\left(b+a\right)}{4\left(b+a\right)}\right]\)
\(A\le\left[\dfrac{b}{4}+\dfrac{a}{4}+\dfrac{c}{4}\right]\)
\(A\le\dfrac{b+a+c}{4}=\dfrac{1}{4}\)
Đẳng thức khi \(a=b=c=\dfrac{1}{3}\)
Xong rồi đó mỏi cái lưng
Áp dụng BĐT Cauchy cho từng cặp số:
\(\dfrac{ab}{c+1}=\dfrac{bc}{a+1}\); \(\dfrac{bc}{a+1}=\dfrac{ca}{b+1}\) ; \(\dfrac{ac}{b+1}=\dfrac{ab}{c+1}\)
Kết quả cuối cùng là \(VT\ge a+b+c=1\)
Dấu " = " xảy ra khi và chỉ khi \(a=b=c=\dfrac{1}{3}\)
Không chắc :v
Ta có:
\(A=\dfrac{ab}{c+1}+\dfrac{bc}{a+1}+\dfrac{ca}{b+1}\)
\(=\dfrac{ab}{c+a+c+b}+\dfrac{bc}{a+b+a+c}+\dfrac{ca}{b+a+b+c}\)
\(\le\dfrac{1}{4}.\left(\dfrac{ab}{c+a}+\dfrac{ab}{b+c}+\dfrac{bc}{a+b}+\dfrac{bc}{c+a}+\dfrac{ca}{a+b}+\dfrac{ca}{b+c}\right)\)
\(=\dfrac{1}{4}\left[\left(\dfrac{ab}{c+a}+\dfrac{bc}{c+a}\right)+\left(\dfrac{ab}{b+c}+\dfrac{ca}{b+c}\right)+\left(\dfrac{bc}{a+b}+\dfrac{ca}{a+b}\right)\right]\)
\(=\dfrac{1}{4}.\left(a+b+c\right)=\dfrac{1}{4}\)
Dấu = xảy ra khi \(a=b=c=\dfrac{1}{3}\)
Các bạn tl giùm đi bài này quyết định số phận của mk ( mk sẽ nói các bạn tick 2 GP cho bạn tl đúng )
Ta có \(A=ab-\dfrac{abc}{c+1}+bc-\dfrac{abc}{a+1}+ac-\dfrac{abc}{b+1}\)
\(=ab+bc+ac-abc\left(\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}\right)\)
Áp dụng BĐT : \(ab+bc+ac\le a^2+b^2+c^2\Rightarrow3\left(ab+bc+ac\right)\le\left(a+b+c\right)^2\)
\(\Rightarrow ab+bc+ac\le\dfrac{\left(a+b+c\right)^2}{3}=\dfrac{1}{3}\) (1)
Áp dụng BDT \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{9}{x+y+z}\)
\(\Rightarrow\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}\ge\dfrac{9}{a+b+c+3}=\dfrac{9}{4}\Rightarrow-\left(\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}\right)\le\dfrac{-9}{4}\)
Áp dụng BDT Cô si : \(\sqrt[3]{abc}\le\dfrac{a+b+c}{3}\Rightarrow abc\le\dfrac{\left(a+b+c\right)^3}{27}=\dfrac{1}{27}\)
\(\Rightarrow-abc\left(\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}\right)\le\dfrac{-9}{4}.\dfrac{1}{27}=\dfrac{-1}{12}\) (2)
Cộng hai vế BDT (1) và (2) ta được
\(ab+bc+ac-abc\left(\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}\right)\le\dfrac{1}{3}-\dfrac{1}{12}=\dfrac{1}{4}\)
\(\Rightarrow A\le\dfrac{1}{4}\Rightarrow MinA=\dfrac{1}{4}\) tại \(x=y=z=\dfrac{1}{3}\)