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áp dụng bất đẳng thức cô-si với 2 số dương.
Ta có
\(a+b\ge2\sqrt{ab}\)
\(b+c\ge2\sqrt{bc}\)
\(c+a\ge2\sqrt{ca}\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8\sqrt{\left(abc\right)^2}=8abc\)(vì a,b,c dương)
![](https://rs.olm.vn/images/avt/0.png?1311)
vì a>0;b>0;c>0\(\Rightarrow\sqrt{a};\sqrt{b};\sqrt{c}\)luôn được xác định
\(\left(\sqrt{a}-\sqrt{b}\right)^2>=0\Rightarrow a-2\sqrt{ab}+b>=0\Rightarrow a+b>=2\sqrt{ab}\)
\(\left(\sqrt{b}-\sqrt{c}\right)^2>=0\Rightarrow b-2\sqrt{bc}+c>=0\Rightarrow b+c>=2\sqrt{bc}\)
\(\left(\sqrt{c}-\sqrt{a}\right)^2>=0\Rightarrow c-2\sqrt{ca}+a>=0\Rightarrow c+a>+2\sqrt{ca}\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)>=2\sqrt{ab}\cdot2\sqrt{bc}\cdot2\sqrt{ca}=8\sqrt{a^2b^2c^2}=8abc\)(đpcm)
dấu = xảy ra khi a=b=c
Áp dụng ĐBT Cauchy - schwarz cho 2 số không âm, ta được:
\(a+b\ge2\sqrt{ab}\)
\(b+c\ge2\sqrt{bc}\)
\(a+c\ge2\sqrt{ac}\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)\ge8\sqrt{\left(abc\right)^2}=8abc\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
khai triển và rút gọn ta được:
\(4a^3+4b^3+4c^3+24abc\ge\left(a+b+c\right)^3.\)<=> \(a^3+b^3+c^3+8abc\ge\left(a+b\right)\left(b+c\right)\left(c+a\right)\)<=> a(a-b)(a-c) + b(b-a)(b-c) +c(c-a)(c-b) +3abc\(\ge0\)
giả sử \(a\ge b\ge c\)
c(c-a)(c-b)\(\ge0\)
a(a-b)(a-c) + b(b-a)(b-c) = (a-b)(a2 - b2 + bc-ac) = (a-b)2(a+b-c) \(\ge0\)
3abc\(\ge0\)
cộng vế theo vế ta được bdt cần chứng minh
dâu '=' khi \(\hept{\begin{cases}c\left(c-a\right)\left(c-b\right)=0\\\left(a-b\right)^2\left(a+b-c\right)=0\\3abc=0\end{cases}}\)=> a=b; c=0
![](https://rs.olm.vn/images/avt/0.png?1311)
Thấy : \(a;b;c\ge0;a+b+c=1\) \(\Rightarrow1-a;1-b;1-c\ge0\)
AD BĐT AM - GM ta được : \(4\left(1-a\right)\left(1-c\right)\le\left(2-a-c\right)^2=\left[2-\left(1-b\right)\right]^2=\left(b+1\right)^2\)
\(\Rightarrow4\left(1-a\right)\left(1-b\right)\left(1-c\right)\le\left(1-b\right)\left(b+1\right)^2=\left(1-b^2\right)\left(b+1\right)\le1.\left(b+1\right)=b+1=b+\left(a+b+c\right)=a+2b+c\)
( đpcm )
![](https://rs.olm.vn/images/avt/0.png?1311)
Xét \(VT=a+2b+c=1+b\left(1\right)\)
Áp dụng BĐT AG-GM:
\(4\left(1-a\right)\left(1-c\right)\le\left(1-a+1-c\right)^2=\left(2-a-c\right)^2=\left(1+a+b+c-a-c\right)^2=\left(1+b\right)^2\left(2\right)\)
\(\Rightarrow4\left(1-a\right)\left(1-b\right)\left(1-c\right)\le\left(1-b\right)\left(1+b\right)^2\)
Mà \(\left(1-b\right)\left(1+b\right)^2-\left(1-b\right)=\left(1+b\right)\left(1-b^2-1\right)=-b^2\left(1+b\right)\le0,\forall b\ge0\)
Do đó \(\left(1-b\right)\left(1+b\right)^2\le1+b\left(3\right)\)
Từ \(\left(1\right)\left(2\right)\left(3\right)\) ta có ĐPCM
Dấu "=" \(\Leftrightarrow a=c=\dfrac{1}{2};b=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=2a+ab\left(2+c\right)\le2a+\dfrac{a}{4}\left(b+2+c\right)^2=2a+\dfrac{a}{4}\left(7-a\right)^2\)
\(P\le\dfrac{1}{4}\left(a^3-14a^2+57a-72\right)+18=18-\dfrac{1}{4}\left(8-a\right)\left(a-3\right)^2\le18\)
\(P_{max}=18\) khi \(\left(a;b;c\right)=\left(3;2;0\right)\)
thầy ơi tại sao dòng một lại có 2a+\(\dfrac{a}{4}\)(b+2+c)^2 ạ?
![](https://rs.olm.vn/images/avt/0.png?1311)
a.
Bình phương 2 vế, BĐT cần chứng minh trở thành:
\(\sqrt{\left(a^2+1\right)\left(b^2+1\right)}+\sqrt{\left(b^2+1\right)\left(c^2+1\right)}+\sqrt{\left(c^2+1\right)\left(a^2+1\right)}\ge6\)
Ta có:
\(\sqrt{\left(a^2+1\right)\left(1+b^2\right)}\ge\sqrt{\left(a+b\right)^2}=a+b\)
Tương tự cộng lại:
\(\sqrt{\left(a^2+1\right)\left(b^2+1\right)}+\sqrt{\left(b^2+1\right)\left(c^2+1\right)}+\sqrt{\left(c^2+1\right)\left(a^2+1\right)}\ge2\left(a+b+c\right)=6\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
b.
\(\sum\dfrac{a+1}{a^2+2a+3}=\sum\dfrac{a+1}{a^2+1+2a+2}\le\sum\dfrac{a+1}{4a+2}\)
Nên ta chỉ cần chứng minh:
\(\sum\dfrac{a+1}{4a+2}\le1\Leftrightarrow\sum\dfrac{4a+4}{4a+2}\le4\)
\(\Leftrightarrow\sum\dfrac{1}{2a+1}\ge1\)
Đúng đo: \(\dfrac{1}{2a+1}+\dfrac{1}{2b+1}+\dfrac{1}{2c+1}\ge\dfrac{9}{2\left(a+b+c\right)+3}=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\Rightarrow A^2=\left(\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\right)^2\)
\(\Rightarrow A^2\le\left(1+1+1\right)\left(\sqrt{a+1}^2+\sqrt{b+1}^2+\sqrt{c+1}^2\right)\left(bunhiacopxki\right)\)
\(\Rightarrow A^2\le\left(1+1+1\right)\left(a+1+b+1+c+1\right)\)
\(\Rightarrow A^2\le3\left(a+b+c+3\right)=3.4=12\Rightarrow A\le\sqrt{12}< 3,5\left(dpcm\right)\)
áp dụng BDT AM-GM
\(=>a+b\ge2\sqrt{ab}\)
\(=>b+c\ge2\sqrt{bc}\)
\(=>c+a\ge2\sqrt{ca}\)
\(=>VT\ge2.2.2\sqrt{ab.bc.ca}=8abc\left(dpcm\right)\)
dấu"=" xảy ra<=>a=b=c
Áp dụng bất đẳng thức AM-GM:
\(a+b\ge2\sqrt{ab}\left(1\right)\\ a+c\ge2\sqrt{ac}\left(2\right)\\ b+c\ge2\sqrt{bc}\left(3\right)\)
Nhân vế theo vế \(\left(1\right)\left(2\right)\left(3\right)\Rightarrow\left(a+b\right)\left(a+c\right)\left(b+c\ge\right)8abc\) ( với \(a,b,c\ge0\) )