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Dễ thấy: \(a^2;b^2;c^2\ge0\forall a;b;c\) mà \(a;b;c\ne0\) nên chỉ có \(a,b,c>0\)
Áp dụng BĐT AM-GM ta có:
\(a^2+\frac{1}{a^2}\ge2\sqrt{a^2\cdot\frac{1}{a^2}}=2\sqrt{1}=2\)
\(b^2+\frac{1}{b^2}\ge2\sqrt{b^2\cdot\frac{1}{b^2}}=2\sqrt{1}=2\)
\(c^2+\frac{1}{c^2}\ge2\sqrt{c^2\cdot\frac{1}{c^2}}=2\sqrt{1}=2\)
Nhân theo vế 3 BĐT trên ta có:
\(\left(a^2+\frac{1}{a^2}\right)\left(b^2+\frac{1}{b^2}\right)\left(c^2+\frac{1}{c^2}\right)\ge2\cdot2\cdot2=8\)
Đẳng thức xảy ra khi \(a=b=c\)
Áp dụng bđt Cauchy Schwarz dạng Engel ta có:
\(\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}\right)\ge\left(a+b+c\right).\dfrac{\left(1+1+1\right)^2}{2\left(a+b+c\right)}\)
\(\ge\dfrac{9}{2}\left(đpcm\right)\)
2.
Từ giả thiết, ta có :
\(\frac{1}{1+a}\ge1-\frac{1}{1+b}+1-\frac{1}{1+c}+1-\frac{1}{1+d}\)
\(=\frac{b}{1+b}+\frac{c}{1+c}+\frac{d}{1+d}\ge3\sqrt[3]{\frac{b.c.d}{\left(1+b\right)\left(1+c\right)\left(1+d\right)}}\)
Tương tự, ta cũng có :
\(\frac{1}{1+b}\ge3\sqrt[3]{\frac{c.d.a}{\left(1+c\right)\left(1+d\right)\left(1+a\right)}}\)
\(\frac{1}{1+c}\ge3\sqrt[3]{\frac{abd}{\left(1+a\right)\left(1+b\right)\left(1+d\right)}}\)
\(\frac{1}{1+d}\ge3\sqrt[3]{\frac{abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)
Nhân vế theo vế 4 BĐT vừa chững minh rồi rút gọn ta được :
\(abcd\le\frac{1}{81}\left(đpcm\right)\)
2) Từ \(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}+\frac{1}{1+d}\ge3.\)
\(\Rightarrow\frac{1}{1+a}\ge\left(1-\frac{1}{1+b}\right)+\left(1-\frac{1}{1+c}\right)+\left(1-\frac{1}{1+d}\right)\)
\(=\frac{b}{1+b}+\frac{c}{1+c}+\frac{d}{1+d}\ge3\sqrt[3]{\frac{bcd}{\left(1+b\right)\left(1+c\right)\left(1+d\right)}}.\)(BĐT AM-GM)
Tương tự :
\(\frac{1}{1+b}\ge3\sqrt[3]{\frac{acd}{\left(1+a\right)\left(1+c\right)\left(1+d\right)}}\)
\(\frac{1}{1+c}\ge3\sqrt[3]{\frac{abd}{\left(1+a\right)\left(1+b\right)\left(1+d\right)}}\)
\(\frac{1}{1+d}\ge3\sqrt[3]{\frac{abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}.\)
Từ đó suy ra:
\(\frac{1}{1+a}.\frac{1}{1+b}.\frac{1}{1+c}.\frac{1}{1+d}\ge3.3.3.3\sqrt[3]{\frac{\left(abcd\right)^3}{\left[\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)\right]^3}}\)
\(\Leftrightarrow\frac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)}\ge\frac{81abcd}{\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)}.\)
\(\Leftrightarrow81abcd\le1\Leftrightarrow abcd\le\frac{1}{81}\)
Dấu '=' xảy ra khi \(a=b=c=d=\frac{1}{3}.\)
3)Ta có: \(\left(\sqrt{a}+\sqrt{b}\right)^8=\left[\left(\sqrt{a}+\sqrt{b}\right)^2\right]^4=\left(a+b+2\sqrt{ab}\right)^4.\)(1)
Với \(a,b\ge0\),áp dụng BĐT AM-GM cho (a+b) và (\(2\sqrt{ab}\)) ta được
\(\left(a+b\right)+2\sqrt{ab}\ge2\sqrt{\left(a+b\right)2\sqrt{ab}}\)(2)
Từ (1) và (2) suy ra:
\(\left(\sqrt{a}+\sqrt{b}\right)^8\ge\left(2\sqrt{\left(a+b\right)2\sqrt{ab}}\right)^4\)
\(\Leftrightarrow\left(\sqrt{a}+\sqrt{b}\right)^8\ge64ab\left(a+b\right)^2.\)
Dấu '=' xảy ra khi \(a+b=2\sqrt{ab}\Leftrightarrow a=b\)
1) Với \(x\le\frac{2}{3}\Rightarrow2-3x\ge0\)
Khi đó ,áp dụng bất đẳng thức AM-GM cho 2 số ta được:
\(\left(2-3x\right)+\frac{9}{2-3x}\ge2\sqrt{\left(2-3x\right)\frac{9}{2-3x}}=2.3=6\)
\(\Leftrightarrow2+\left(2-3x\right)+\frac{9}{2-3x}\ge2+6\)
\(\Leftrightarrow4-3x+\frac{9}{2-3x}\ge8\)
Dấu '=' xảy ra khi \(2-3x=\frac{9}{2-3x}\Leftrightarrow\left(2-3x\right)^2=9\Leftrightarrow2-3x=3\Leftrightarrow x=-\frac{1}{3}\)( vì 2-3x>0)
\(A=\dfrac{\left(a-b\right)^2}{ab}+\dfrac{\left(b-c\right)^2}{bc}+\dfrac{\left(c-a\right)^2}{ca}\)
\(B=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
..................................
\(A=\dfrac{a^2+b^2-2ab}{ab}+\dfrac{b^2-2ab+c^2}{bc}+c^2+a^2-\dfrac{2ca}{ca}\)
\(A=\left(\dfrac{a}{b}+\dfrac{b}{a}-2\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}-2\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}-2\right)=\dfrac{\left(b+c\right)}{a}+\dfrac{a+c}{b}+\dfrac{a+b}{c}-6\)
\(A=\left[\dfrac{\left(b+c\right)}{a}+1\right]+\left[\dfrac{\left(a+c\right)}{b}+1\right]+\left[\dfrac{\left(a+b\right)}{c}+1\right]-9\)
\(A=\dfrac{\left(a+b+c\right)}{a}+\dfrac{\left(a+b+c\right)}{b}+\left[\dfrac{\left(a+b+c\right)}{c}\right]-9\)
\(A=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)-9\)
Ket luan
\(A\ne B\) => đề sai--> hoặc mình công trừ sai
\(\Leftrightarrow a^2\left(\dfrac{1}{a+b}-\dfrac{1}{b+c}\right)+b^2\left(\dfrac{1}{b+c}-\dfrac{1}{c+a}\right)+c^2\left(\dfrac{1}{c+a}-\dfrac{1}{a+b}\right)=0\)
\(\Leftrightarrow\dfrac{a^2\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}+\dfrac{b^2\left(a-b\right)}{\left(a+c\right)\left(b+c\right)}+\dfrac{c^2\left(b-c\right)}{\left(a+b\right)\left(a+c\right)}=0\)
\(\Leftrightarrow a^2\left(c-a\right)\left(c+a\right)+b^2\left(a-b\right)\left(a+b\right)+c^2\left(b-c\right)\left(b+c\right)=0\)
\(\Leftrightarrow a^2\left(c^2-a^2\right)+b^2\left(a^2-b^2\right)+c^2\left(b^2-c^2\right)=0\)
\(\Leftrightarrow a^2c^2+a^2b^2+b^2c^2-a^4-b^4-c^4=0\)
\(\Leftrightarrow2a^4+2b^4+2c^4-2a^2b^2-2a^2c^2-2b^2c^2=0\)
\(\Leftrightarrow\left(a^2-b^2\right)^2+\left(a^2-c^2\right)^2+\left(b^2-c^2\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}a^2-b^2=0\\a^2-c^2=0\\b^2-c^2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(a-b\right)\left(a+b\right)=0\\\left(a-c\right)\left(a+c\right)=0\\\left(b-c\right)\left(b+c\right)=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a-b=0\\a-c=0\\b-c=0\end{matrix}\right.\) (do \(\left(a+b\right)\left(a+c\right)\left(b+c\right)\ne0\) \(\Rightarrow\left\{{}\begin{matrix}a+b\ne0\\a+c\ne0\\b+c\ne0\end{matrix}\right.\))
\(\Rightarrow a=b=c\)
\(\dfrac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}\)
\(=\dfrac{a^2+\left(a-c\right)^2+c^2+2\left(ab-ac-bc\right)}{b^2+\left(b-c\right)^2+c^2+2\left(ab-ac-bc\right)}\)
\(=\dfrac{a^2+a^2-2ac+c^2+c^2+2ab-2ac-2bc}{b^2+b^2-2bc+c^2+c^2+2ab-2ac-2bc}\)
\(=\dfrac{2a^2+2c^2-4ac+2ab-2bc}{2b^2+2c^2-4bc+2ab-2ac}\)
\(=\dfrac{\left(a-c\right)^2+b\left(a-c\right)}{\left(b-c\right)^2+a\left(b-c\right)}\)
\(=\dfrac{\left(a-c\right)\left(a-c+b\right)}{\left(b-c\right)\left(a-c+b\right)}=\dfrac{a-c}{b-c}\left(đpcm\right)\)
\((\dfrac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}+\dfrac{1}{\left(c-a\right)\left(b^2+ba-c^2-ca\right)}+\dfrac{1}{\left(a-b\right)\left(c^2+cb-a^2-ab\right)}=0 \)
\(\Leftrightarrow\dfrac{1}{\left(b-c\right)\left[\left(a-b\right)\left(a+b\right)+c\left(a-b\right)\right]}+\dfrac{1}{\left(c-a\right)\left[\left(b-c\right)\left(b+c\right)+a\left(b-c\right)\right]}+\dfrac{1}{\left(a-b\right)\left[\left(c-a\right)\left(c+a\right)+b\left(c-a\right)\right]}=0\)
\(\Leftrightarrow\dfrac{1}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}+\dfrac{1}{\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}+\dfrac{1}{\left(a-b\right)\left(c-a\right)\left(a+b+c\right)}=0\)
\(\Leftrightarrow\dfrac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)}=0\)
\(\Leftrightarrow\dfrac{0}{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)}=0\)(t/m)
Suy ra ta được Đt cần chứng minh.
Chúc bạn học tốt với hoc24 nha
Lời giải:
Ta có:
\(\frac{1}{(b-c)(a^2+ac-b^2-bc)}+\frac{1}{(c-a)(b^2+bc-c^2-ca)}+\frac{1}{(a-b)(c^2+cb-a^2-ab)}\)
\(=\frac{1}{(b-c)[(a^2-b^2)+(ac-bc)]}+\frac{1}{(c-a)[(b^2-c^2)+(ba-ca)]}+\frac{1}{(a-b)[(c^2-a^2)+(cb-ab)]}\)
\(=\frac{1}{(b-c)[(a-b)(a+b)+c(a-b)]}+\frac{1}{(c-a)[(b-c)(b+c)+a(b-c)]}+\frac{1}{(a-b)[(c-a)(c+a)+b(c-a)]}\)
\(=\frac{1}{(b-c)(a-b)(a+b+c)}+\frac{1}{(c-a)(b-c)(b+c+a)}+\frac{1}{(a-b)(c-a)(c+a+b)}\)
\(=\frac{(c-a)+(a-b)+(b-c)}{(a-b)(b-c)(c-a)(a+b+c)}=\frac{0}{(a-b)(b-c)(c-a)(a+b+c)}=0\)
Ta có đpcm.
Lời giải
\(\left(a^2+\dfrac{1}{a^2}\right)\left(b^2+\dfrac{1}{b^2}\right)\left(c^2+\dfrac{1}{c^2}\right)\ge8\)
\(A=\left(a^2+\dfrac{1}{a^2}\right)\left(b^2+\dfrac{1}{b^2}\right)\left(c^2+\dfrac{1}{c^2}\right)\)
\(A=\left[\left(a^2+\dfrac{1}{a^2}-2\right)+2\right].\left[\left(a^2+\dfrac{1}{a^2}-2\right)+2\right].\left[\left(a^2+\dfrac{1}{a^2}-2\right)+2\right]\)
\(A=\left[\left(a-\dfrac{1}{a}\right)^2+2\right].\left[\left(a-\dfrac{1}{a}\right)^2+2\right].\left[\left(a-\dfrac{1}{a}\right)^2+2\right]\)Thừa nhận cần c/m câu khác: \(\left(x-\dfrac{1}{x}\right)^2\ge0\forall x\ne0\)
\(\Rightarrow A\ge\left[\left(0\right)+2\right].\left[\left(0\right)+2\right].\left[\left(0\right)+2\right]=8\)
\(\Rightarrow A\ge8\forall_{a,b,c\ne0}\)=> dpcm
Đẳng thức khi \(\left\{{}\begin{matrix}\left|a\right|=1\\\left|b\right|=1\\\left|c\right|=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\pm1\\b=\pm1\\c=\pm1\end{matrix}\right.\) Không tin bạn thử a=b=c=-1<0 vào thử xem
Có một chút vần đề nha ĐK phải là a,b,c > 0 nhé
bài này ta sẽ chứng minh lần lượt \(a^2+\dfrac{1}{a^2};b^2+\dfrac{1}{b^2};c^2+\dfrac{1}{c^2}\)lớn hơn hoặc bằng 2
Ta sẽ giả sử
\(a^2+\dfrac{1}{a^2}\ge2\)(2)
\(\Leftrightarrow a^2-2+\dfrac{1}{a^2}\ge0\Leftrightarrow a^2-2a\times\dfrac{1}{a}+\dfrac{1}{a^2}\ge0\)
\(\Leftrightarrow\left(a-\dfrac{1}{a}\right)^2\ge0\)(luôn đúng) (1)
BĐT (2) đúng suy ra BĐT (1) đúng
Dấu '=' xảy ra khi và chỉ khi \(a=\dfrac{1}{a}\Leftrightarrow a^2=1\Leftrightarrow a=1\)(*)
CMTT ta có : \(b^2+\dfrac{1}{b^2}\ge2\) (=) b = 1 (**)
\(c^2+\dfrac{1}{c^2}\ge2\) (=) c = 1 (***)
Nhân vế theo vế của (*) , (**) , (***) ta được
\(\left(a^2+\dfrac{1}{a^2}\right).\left(b^2+\dfrac{1}{b^2}\right).\left(c^2+\dfrac{1}{c^2}\right)\ge2^3=8\)(đpcm)
Dấu "=" xảy ra khi và chỉ khi a = b = c = 1