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Đề bài sai, phản ví dụ: \(a=b=0,c=1\)
BĐT này chỉ đúng khi a;b;c là độ dài 3 cạnh của 1 tam giác
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1 :
a) Ta có : \(\left(1-a\right)\left(1-b\right)\left(1-c\right)=\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Áp dụng bđt Cauchy : \(a+b\ge2\sqrt{ab}\) , \(b+c\ge2\sqrt{bc}\) , \(c+a\ge2\sqrt{ca}\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\) hay \(\left(1-a\right)\left(1-b\right)\left(1-c\right)\ge8abc\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bunhiacopxki:
\(\left(a^2+bc+ca\right)\left(b^2+bc+ca\right)\ge\left(ab+bc+ca\right)^2\)
\(\Rightarrow\dfrac{ab}{a^2+bc+ca}\le\dfrac{ab\left(b^2+bc+ca\right)}{\left(ab+bc+ca\right)^2}\)
Tương tự: \(\dfrac{bc}{b^2+ca+ab}\le\dfrac{bc\left(c^2+ca+ab\right)}{\left(ab+bc+ca\right)^2}\)
\(\dfrac{ca}{c^2+ab+bc}\le\dfrac{ca\left(a^2+ab+bc\right)}{\left(ab+bc+ca\right)^2}\)
\(\Rightarrow VT\le\dfrac{ab\left(b^2+bc+ca\right)+bc\left(c^2+ca+ab\right)+ca\left(a^2+ab+bc\right)}{\left(ab+bc+ca\right)^2}\)
Nên ta chỉ cần chứng minh:
\(\dfrac{ab\left(b^2+bc+ca\right)+bc\left(c^2+ca+ab\right)+ca\left(a^2+ab+bc\right)}{\left(ab+bc+ca\right)^2}\le\dfrac{a^2+c^2+c^2}{ab+bc+ca}\)
\(\Leftrightarrow ab\left(b^2+bc+ca\right)+bc\left(c^2+ca+ab\right)+ca\left(a^2+ab+bc\right)\le\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)\)
Nhân phá và rút gọn 2 vế:
\(\Leftrightarrow a^3b+b^3c+c^3a\ge abc\left(a+b+c\right)\)
\(\Leftrightarrow\dfrac{a^3b+b^3c+c^3a}{abc}\ge a+b+c\)
\(\Leftrightarrow\dfrac{a^2}{c}+\dfrac{b^2}{a}+\dfrac{c^2}{b}\ge a+b+c\)
Đúng do: \(\dfrac{a^2}{c}+\dfrac{b^2}{a}+\dfrac{c^2}{b}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c}=a+b+c\)
Dấu "=" xảy ra khi \(a=b=c\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Không mất tính tổng quát giả sử \(a\ge b\ge c>0\Rightarrow\hept{\begin{cases}b+c\le a+c\le a+b\\\frac{a^a}{b+c}\ge\frac{b^a}{c+a}\ge\frac{c^a}{a+b}\end{cases}}\)
Sử dụng bất đẳng thức Chebyshev cho 2 dãy đơn ngược chiều ta có:
\(VT\left(1\right)=\frac{1}{2\left(a+b+c\right)}\left(\frac{a^a}{b+c}+\frac{b^a}{c+a}+\frac{c^a}{a+b}\right)\left[\left(b+c\right)+\left(c+a\right)+\left(a+b\right)\right]\ge\)
\(\frac{1}{2\left(a+b+c\right)}\cdot3\left[\frac{a^a}{b+c}\left(b+c\right)+\frac{b^a}{c+a}\left(c+a\right)+\frac{c^a}{a+b}\left(a+b\right)\right]=\frac{3\left(a^a+b^a+c^a\right)}{2\left(a+b+c\right)}\)\(=\frac{3}{2}\cdot\frac{a^a+b^a+c^a}{a+b+c}\)
=> đpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
Làm đại luôn mặc dù chưa xong xD. Có sai sót gì cho xin lỗi nha!
Đặt: \(M=\frac{a^2+bc}{\left(b+c\right)^2}+\frac{b^2+ca}{\left(c+a\right)^2}+\frac{c^2+ab}{\left(a+b\right)^2}\)
\(M=\frac{\frac{1}{\left(b+c\right)^2}}{\frac{1}{a^2+bc}}+\frac{\frac{1}{\left(c+a\right)^2}}{\frac{1}{b^2+ca}}+\frac{\frac{1}{\left(a+b\right)^2}}{\frac{1}{c^2+ab}}\)
Áp dụng Bđt AM-GM dạng Engel:
\(M\ge\frac{\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)^2}{\frac{1}{a^2+bc}+\frac{1}{b^2+ca}+\frac{1}{c^2+ab}}\)
Chuẩn hóa: \(a+b+c=3\)
Có: \(A=\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)^2\ge\left(\frac{9}{2\left(a+b+c\right)}\right)^2=\left(\frac{3}{2}\right)^2\)
CM:\(B=\frac{1}{a^2+bc}+\frac{1}{b^2+ca}+\frac{1}{c^2+ab}\le\frac{3}{2}\)so what ? Tới đây k biết làm.
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=\frac{1}{a^2+2b^2+3}+\frac{1}{b^2+2c^2+3}+\frac{1}{c^2+2a^2+3}\)
Áp dụng bất đẳng thức cô-si, ta có:
\(a^2+b^2\ge2.\sqrt{a^2.b^2}=>a^2+b^2\ge2ab\)
\(b^2+1\ge2.\sqrt{b^2.1}=>b^2+1\ge2b\)
=>\(a^2+b^2+b^2+1\ge2ab+2b\)
=>\(a^2+2b^2+1+2\ge2ab+2b+2\)
=>\(a^2+2b^2+3\ge2ab+2b+2\)
=>\(a^2+2b^2+3\ge2\left(ab+b+1\right)\)
=>\(\frac{1}{a^2+2b^2+3}\le\frac{1}{2.\left(ab+b+1\right)}\)
Chứng minh tương tự, ta có:
\(\frac{1}{b^2+2c^2+3}\le\frac{1}{2.\left(bc+c+1\right)}\)
\(\frac{1}{c^2+2a^2+3}\le\frac{1}{2.\left(ca+a+1\right)}\)
=>\(\frac{1}{a^2+2b^2+3}+\frac{1}{b^2+2c^2+3}+\frac{1}{c^2+2a^2+3}\le\frac{1}{2.\left(ab+b+1\right)}+\frac{1}{2.\left(bc+c+1\right)}+\frac{1}{2.\left(ca+a+1\right)}\)
=>\(A\le\frac{1}{2}.\frac{1}{ab+b+1}+\frac{1}{2}.\frac{1}{bc+c+1}+\frac{1}{2}.\frac{1}{ca+a+1}\)
=>\(A\le\frac{1}{2}.\left(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}\right)\)
=>\(A\le\frac{1}{2}.\left(\frac{ca}{ca.\left(ab+b+1\right)}+\frac{a}{a.\left(bc+c+1\right)}+\frac{1}{ca+a+1}\right)\)
=>\(A\le\frac{1}{2}.\left(\frac{ca}{abc.c+abc+ca}+\frac{a}{abc+ca+a}+\frac{1}{ca+a+1}\right)\)
Vì abc=1(theo giả thiết)
=>\(A\le\frac{1}{2}.\left(\frac{ca}{c+1+ca}+\frac{a}{1+ca+a}+\frac{1}{ca+a+1}\right)\)
=>\(A\le\frac{1}{2}.\left(\frac{ca}{ca+a+1}+\frac{a}{ca+a+1}+\frac{1}{ca+a+1}\right)\)
=>\(A\le\frac{1}{2}.\frac{ca+a+1}{ca+a+1}\)
=>\(A\le\frac{1}{2}.1\)
=>\(A\le\frac{1}{2}\)
=>\(\frac{1}{a^2+2b^2+3}+\frac{1}{b^2+2c^2+3}+\frac{1}{c^2+2a^2+3}\le\frac{1}{2}\)
=>ĐPCM
![](https://rs.olm.vn/images/avt/0.png?1311)
Định đi ngủ mà chợt nhớ lúc chiều có hứa là làm giúp chủ tus nên h phải làm =)))
1. Giải
Ta chứng minh với mọi x, y luôn có : \(\frac{x+y}{2}\cdot\frac{x^3+y^3}{2}\le\frac{x^4+y^4}{2}\) (1)
\(\Rightarrow\left(1\right)\Leftrightarrow\left(x+y\right)\left(x^3+y^3\right)\le2\left(x^4+y^4\right)\)
\(\Leftrightarrow xy\left(x^2+y^2\right)\le x^4+y^4\)
\(\Leftrightarrow\left(x-y\right)^2\left[\left(\frac{x+y}{2}\right)^2+\frac{3y^2}{4}\right]\ge0\)
ÁP DỤNG (1) ta được
\(\frac{a+b}{2}\cdot\frac{a^2+b^2}{2}\cdot\frac{a^3+b^3}{2}=\left[\frac{a+b}{2}\cdot\frac{a^3+b^3}{2}\right]\cdot\frac{a^2+b^2}{2}\)
\(\Leftrightarrow\left[\frac{a+b}{2}\cdot\frac{a^3+b^3}{2}\right]\cdot\frac{a^2+b^2}{2}\le\frac{a^4+b^4}{2}\cdot\frac{a^2+b^2}{2}\le\frac{a^6+b^6}{2}\left(đpcm\right)\)
2. Ta biến đổi các Đẳng thức : \(a^2+b^2+c^2-\left(ab+bc+ca\right)\ge0\)
\(\Leftrightarrow\left(\frac{a^2}{2}-ab+\frac{b^2}{2}\right)+\left(\frac{b^2}{2}-bc+\frac{c^2}{2}\right)-\left(\frac{c^2}{2}-ca+\frac{a^2}{2}\right)\ge0\)
\(\Leftrightarrow\left(\frac{a}{\sqrt{2}}-\frac{b}{\sqrt{2}}\right)^2+\left(\frac{b}{\sqrt{2}}-\frac{c}{\sqrt{2}}\right)+\left(\frac{c}{\sqrt{2}}-\frac{a}{\sqrt{2}}\right)\ge0\left(đpcm\right)\)