\(a\left(b-1\right)+b\left(1-c\right)+c\left(1-a\right)\le1\\ \Leftrightarrow-abc+ab+bc+ca-a-b-c+1\le2-abc\\ \Leftrightarrow\left(1-a\right)\left(1-b\right)\left(1-c\right)\le2-abc\)

lại có \(abc\le1\) nên \(2-abc\ge1\)

ta chứng minh \(\left(1-a\right)\left(1-b\right)\left(1-c\right)\le1\)

luôn đúng do \(0\le a;b;c\le1\)

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Không mất tính tổng quát, giả sử \(a\ge b\ge c\).

Khi đó: \(\left(a-b\right)\left(b-c\right)\ge0\)

\(\Leftrightarrow ab+bc\ge ac+b^2\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{c}+1\ge\dfrac{a}{b}+\dfrac{b}{c}\\\dfrac{c}{a}+1\ge\dfrac{c}{b}+\dfrac{b}{a}\end{matrix}\right.\)

\(\Rightarrow\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}\le2+2\left(\dfrac{a}{c}+\dfrac{c}{a}\right)\)

Vì \(1\le c\le a\le2\Rightarrow\left(\dfrac{a}{c}-2\right)\left(\dfrac{2a}{c}-1\right)\le0\)

\(\Leftrightarrow\dfrac{a}{c}+\dfrac{c}{a}\le\dfrac{5}{2}\)

\(\Rightarrow\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}\le7\)

\(\Leftrightarrow\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\le10\)

Đẳng thức xảy ra khi \(a=b=2;c=1\) và các hoán vị.

Không mất tính tổng quát, giả sử \(a\ge b\ge c\)

\(\Rightarrow P\le\dfrac{a}{b+c+1}+\dfrac{b}{b+c+1}+\dfrac{c}{b+c+1}+\left(1-a\right)\left(1-b\right)\left(1-c\right)\)

\(\Rightarrow P\le\dfrac{a+b+c}{b+c+1}+\left(1-a\right)\left(1-b\right)\left(1-c\right)=\dfrac{a-1}{b+c+1}+\left(1-a\right)\left(1-b\right)\left(1-c\right)+1\)

\(\Rightarrow P\le\left(1-a\right)\left[\left(1-b\right)\left(1-c\right)-\dfrac{1}{b+c+1}\right]+1\le\left(1-a\right)\left[\left(1-b\right)\left(1-c\right)-\dfrac{1}{bc+b+c+1}\right]+1\)

\(\Rightarrow P\le\left(1-a\right)\left[\left(1-b\right)\left(1-c\right)-\dfrac{1}{\left(1+b\right)\left(1+c\right)}\right]+1\)

\(\Rightarrow P\le\left(1-a\right)\left(\dfrac{\left(1-b^2\right)\left(1-c^2\right)-1}{\left(1+b\right)\left(1+c\right)}\right)+1\)

Do \(a;b;c\le1\Rightarrow\left\{{}\begin{matrix}1-a\ge0\\\left(1-b^2\right)\left(1-c^2\right)\le1\\\end{matrix}\right.\) \(\Rightarrow\left(1-a\right)\left[\dfrac{\left(1-b^2\right)\left(1-c^2\right)-1}{\left(1+b\right)\left(1+c\right)}\right]\le0\)

\(\Rightarrow P\le1\)

\(P_{max}=1\) khi \(\left(a;b;c\right)=\left(0;0;0\right);\left(1;1;1\right);\left(0;1;1\right);\left(0;0;1\right)\) và các hoán vị

Không mất tính tổng quát ta giả sử \(0\le a\le b\le c\le1\)

\(\Rightarrow\left(1-c\right)\left(b-a\right)\ge0\)\(\Leftrightarrow b-a-bc+ac\ge0\Leftrightarrow ac+b\ge a+bc\)

\(\Leftrightarrow ac+b+1\ge a+bc+1\)\(\Rightarrow\dfrac{a}{ac+b+1}\le\dfrac{a}{a+bc+1}\)(1)

ta cũng có : \(\left(1-b\right)\left(c-a\right)\ge0\Leftrightarrow ab+c\ge a+bc\Leftrightarrow ab+c+1\ge a+bc+1\)

\(\Rightarrow\dfrac{b}{ab+c+1}\le\dfrac{b}{a+bc+1}\) mà \(b\le c\le1\)

nên \(\dfrac{b}{a+bc+1}\le\dfrac{bc}{a+bc+1}\) \(\Rightarrow\dfrac{b}{ab+c+1}\le\dfrac{bc}{a+bc+1}\)(2)

ta lại có : \(\dfrac{c}{a+bc+1}\le\dfrac{1}{a+bc+1}\)(3)

Cộng Ba vế BĐT (1) (2) (3) lại với nhau ta có

\(\dfrac{a}{1+b+ac}+\dfrac{b}{1+c+ab}+\dfrac{c}{1+a+bc}\le\dfrac{a+bc+1}{a+bc+1}=1\)

không cần giả sử gì hết , phang luôn \(\left(a-1\right)\left(b-1\right)\ge0\) (:V)

\(\Leftrightarrow ab+1\ge a+b\Leftrightarrow ab+c+1\ge a+b+c\)

\(\Rightarrow VT\le\sum\dfrac{b}{a+b+c}=1\)

Dấu = xảy ra : 2 số bằng 1 , số còn lại tùy ý

Mở rộng : \(\forall a,b,c\in\left[0;1\right]\).Cmr:

\(\dfrac{a}{b+c+1}+\dfrac{b}{c+a+1}+\dfrac{c}{a+b+1}+\left(1-a\right)\left(1-b\right)\left(1-c\right)\le1\)

( Olympic USA 1980 )

yến nhi hỏi bài à

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Giúp mik ik, mik cần gấp lắm!!!

1 đường thẳng a cắt 2 đoạn thẳng

2 có 4 đoạn thẳng cắt a

Do vai trò a;b;c như nhau, không mất tính tổng quát giả sử \(2\ge a\ge b\ge c\ge1\) 

\(\Rightarrow1\le\dfrac{a}{c}\le2\)

Đồng thời \(\Rightarrow\left(a-b\right)\left(b-c\right)\ge0\Leftrightarrow ab+bc\ge b^2+ac\) (1)

Chia 2 vế của (1) cho \(bc:\)

\(\Rightarrow\dfrac{a}{c}+1\ge\dfrac{b}{c}+\dfrac{a}{b}\)

Chia 2 vế của (1) cho \(ab\Rightarrow1+\dfrac{c}{a}\ge\dfrac{b}{a}+\dfrac{c}{b}\)

Cộng vế: \(\Rightarrow\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{b}{c}+\dfrac{c}{b}\le\dfrac{a}{c}+\dfrac{c}{a}+2\)

Do đó:

\(S=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\left(\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{b}{c}+\dfrac{c}{b}\right)+\dfrac{a}{c}+\dfrac{c}{a}+3\)

\(S\le2\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+5\)

Đặt \(\dfrac{a}{c}=x\Rightarrow1\le x\le2\)

\(S\le2\left(x+\dfrac{1}{x}\right)+5=\dfrac{2x^2-5x+2}{x}+10=\dfrac{\left(2x-1\right)\left(x-2\right)}{x}+10\le10\)

\(S_{max}=10\) khi \(\left(a;b;c\right)=\left(1;1;2\right);\left(1;2;2\right)\) và các hoán vị