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Dùng phương pháp biến đổi tương đương nhé!!!
Ta có : \(\dfrac{1}{1+a^2}\) + \(\dfrac{1}{1+b^2}\) \(\ge\) \(\dfrac{2}{1+ab}\)
<=>( \(\dfrac{1}{1+a^2}\) - \(\dfrac{1}{1+ab}\) ) + ( \(\dfrac{1}{1+b^2}\) - \(\dfrac{1}{1+ab}\) ) \(\ge\) 0
<=> \(\dfrac{1+ab-1-a^2}{\left(1+a^2\right)\left(1+ab\right)}\) + \(\dfrac{1+ab-1-b^2}{\left(1+b^2\right)\left(1+ab\right)}\) \(\ge\) 0
<=> \(\dfrac{ab-a^2}{\left(1+a^2\right)\left(1+ab\right)}\) + \(\dfrac{ab-b^2}{\left(1+b^2\right)\left(1+ab\right)}\) \(\ge\) 0
<=> \(\dfrac{a\left(b-a\right)\left(1+b^2\right)+b\left(a-b\right)\left(1+a^2\right)}{\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)}\) \(\ge\) 0
<=> \(a\left(b-a\right)\left(1+b^2\right)-b\left(b-a\right)\left(1+a^2\right)\) \(\ge\) 0
<=> \(\left(b-a\right)\left(a+ab^2-b-a^2b\right)\) \(\ge\) 0
<=> \(\left(b-a\right)\left[ab\left(b-a\right)-\left(b-a\right)\right]\) \(\ge\) 0
<=> \(\left(b-a\right)\left(b-a\right)\left(ab-1\right)\) \(\ge\) 0
<=> \(\left(b-a\right)^2\left(ab-1\right)\) \(\ge\) 0 (1)
Mà \(\left\{{}\begin{matrix}\left(b-a\right)^2\ge0\\ab-1\ge0\end{matrix}\right.\) ( vì ab \(\ge\)1)
=> \(\left(b-a\right)^2\left(ab-1\right)\) \(\ge\) 0
=> (1) luôn đúng
Vậy đpcm ....
Ta có: \(\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}\ge\dfrac{2}{1+ab}\)
\(\Leftrightarrow\left(\dfrac{1}{1+a^2}-\dfrac{1}{1+b^2}\right)+\left(\dfrac{1}{1+b^2}-\dfrac{1}{1+ab}\right)\ge0\)
\(\Leftrightarrow\dfrac{ab-a^2}{\left(1+a^2\right)\left(1+ab\right)}+\dfrac{ab-b^2}{\left(1+b^2\right)\left(1+ab\right)}\ge0\)
\(\Leftrightarrow\dfrac{a\left(b-a\right)}{\left(1+a^2\right)\left(1+ab\right)}+\dfrac{b\left(a-b\right)}{\left(1+b^2\right)\left(1+ab\right)}\ge0\)
\(\Leftrightarrow\dfrac{\left(b-a\right)^2\left(ab-1\right)}{\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)}\ge0\)
BĐT cuối cùng đúng vì \(a.b\ge1\Rightarrowđpcm\)
a)Svac-so:
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{b+c+c+a+a+b}=\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2\left(đpcm\right)}\)
b)\(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}\ge\dfrac{2}{ab+1}\)
\(\Leftrightarrow\dfrac{1}{a^2+1}-\dfrac{1}{ab+1}+\dfrac{1}{b^2+1}-\dfrac{1}{ab+1}\ge0\)
\(\Leftrightarrow\dfrac{ab+1-a^2-1}{\left(a^2+1\right)\left(ab+1\right)}+\dfrac{ab+1-b^2-1}{\left(b^2+1\right)\left(ab+1\right)}\ge0\)
\(\Leftrightarrow\dfrac{a\left(b-a\right)}{\left(a^2+1\right)\left(ab+1\right)}+\dfrac{b\left(a-b\right)}{\left(b^2+1\right)\left(ab+1\right)}\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(\dfrac{b}{\left(b^2+1\right)\left(ab+1\right)}-\dfrac{a}{\left(a^2+1\right)\left(ab+1\right)}\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(\dfrac{b\left(a^2+1\right)-a\left(b^2+1\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(\dfrac{a^2b+b-ab^2-a}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(\dfrac{ab\left(a-b\right)-\left(a-b\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\cdot\dfrac{ab-1}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\ge0\)(luôn đúng)
Lời giải:
Đề bài phải sửa lại là \(\frac{1}{a^2+1}+\frac{1}{b^2+1}\geq \frac{2}{ab+1}\) em nhé.
Sử dụng pp biến đổi tương đương. Ta có:
\(\frac{1}{a^2+1}+\frac{1}{b^2+1}\geq \frac{2}{ab+1}\)
\(\Leftrightarrow \frac{b^2+1+a^2+1}{(a^2+1)(b^2+1)}\geq \frac{2}{ab+1}\)
\(\Leftrightarrow (ab+1)(a^2+b^2+2)\geq 2(a^2b^2+a^2+b^2+1)\)
\(\Leftrightarrow ab(a^2+b^2)+2ab\geq 2a^2b^2+a^2+b^2\)
\(\Leftrightarrow ab(a^2+b^2-2ab)+2ab-a^2-b^2\geq 0\)
\(\Leftrightarrow ab(a-b)^2-(a-b)^2\geq 0\)
\(\Leftrightarrow (ab-1)(a-b)^2\geq 0\)
BĐT trên luôn đúng vì \(a,b\geq 1\rightarrow ab-1\geq 0\) và \((a-b)^2\geq 0\) )
Ta có đpcm.
Dấu bằng xảy ra khi \(a=b\) hoặc \(ab=1\)
áp dụng bất đằng thức buinhia
\(\left(a+b\right)^2\le2\left(a^2+b^2\right)\Leftrightarrow1\le2\left(a^2+b^2\right)\Rightarrow a^2+b^2\ge\frac{1}{2}\)
\(\left(a^2+b^2\right)^2\le\left(\left(a^2\right)^2+\left(b^2\right)^2\right)2\Leftrightarrow\left(\frac{1}{2}\right)^2\le2\left(a^4+b^4\right)\Rightarrow a^4+b^4\ge\frac{1}{8}\)
bài cuối tương tự
a, \(a^2+b^2\ge\frac{1}{2}\)
Với mọi a, b ta có:
\(\left(a-b\right)^2\ge0\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\)
\(\Leftrightarrow a^2+b^2\ge2ab\)
\(\Leftrightarrow2\left(a^2+b^2\right)\ge a^2+2ab+b^2\)
\(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)
Mà a + b = 1 \(\Rightarrow2\left(a^2+b^2\right)\ge1\)
\(\Leftrightarrow a^2+b^2\ge\frac{1}{2}\)
Vậy \(a^2+b^2\ge\frac{1}{2}\)( đpcm )
Các câu b, c tương tự
Áp dụng bđt Cauchy-Schwarz:
\(a^2+b^2\ge\dfrac{\left(a+b\right)^2}{2}=\dfrac{1}{2}\)
\(a^4+b^4\ge\dfrac{\left(a^2+b^2\right)^2}{2}\ge\dfrac{\left(\dfrac{1}{2}\right)^2}{2}=\dfrac{1}{8}\)
\(a^8+b^8\ge\dfrac{\left(a^4+b^4\right)^2}{2}\ge\dfrac{\left(\dfrac{1}{8}\right)^2}{2}=\dfrac{1}{128}\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\left(1^2+1^2+1^2\right)\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2=9^2\)
\(\Rightarrow3\left(a^2+b^2+c^2\right)\ge9\Rightarrow a^2+b^2+c^2\ge3\)
Lại có: \(a^2+b^2+c^2\ge ab+bc+ac\forall a,b,c\)
\(\Rightarrow3\ge ab+bc+ac\Rightarrow ab+bc+ac\le3\)
Bất đẳng thức ban đầu tương đương với:
\(\dfrac{a^2}{a\left(b^2+1\right)}+\dfrac{b^2}{b\left(c^2+1\right)}+\dfrac{c^2}{c\left(a^2+1\right)}\ge\dfrac{3}{2}\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(\dfrac{a^2}{a\left(b^2+1\right)}+\dfrac{b^2}{b\left(c^2+1\right)}+\dfrac{c^2}{c\left(a^2+1\right)}\ge\dfrac{\left(a+b+c\right)^2}{a\left(b^2+1\right)+b\left(c^2+1\right)+c\left(a^2+1\right)}\)
Áp dụng BĐT AM-GM ta có:
\(\left\{{}\begin{matrix}a\left(b^2+1\right)\ge a\cdot2\sqrt{b^2}=2ba\\b\left(c^2+1\right)\ge b\cdot2\sqrt{c^2}=2cb\\c\left(a^2+1\right)\ge c\cdot2\sqrt{a^2}=2ac\end{matrix}\right.\)
\(\Rightarrow\dfrac{a^2}{a\left(b^2+1\right)}+\dfrac{b^2}{b\left(c^2+1\right)}+\dfrac{c^2}{c\left(a^2+1\right)}\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\)
Mà \(ab+bc+ca\le3\)\(\Rightarrow\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\ge\dfrac{\left(a+b+c\right)^2}{2\cdot3}=\dfrac{9}{6}=\dfrac{3}{2}\)
Đẳng thức xảy ra khi \(a=b=c=1\)
\(VT=\dfrac{a}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{c}{a^2+1}\)
\(VT=a-\dfrac{ab^2}{b^2+1}+b-\dfrac{bc^2}{c^2+1}+c-\dfrac{ca^2}{a^2+1}\)
\(VT=3-\left(\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\right)\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}b^2+1\ge2\sqrt{b^2}=2b\\c^2+1\ge2\sqrt{c^2}=2c\\a^2+1\ge2\sqrt{a^2}=2a\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{ab^2}{b^2+1}\le\dfrac{ab^2}{2b}=\dfrac{ab}{2}\\\dfrac{bc^2}{c^2+1}\le\dfrac{bc^2}{2c}=\dfrac{bc}{2}\\\dfrac{ca^2}{a^2+1}\le\dfrac{ca^2}{2a}=\dfrac{ca}{2}\end{matrix}\right.\)
\(\Rightarrow\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\le\dfrac{ab+bc+ca}{2}\)
\(\Rightarrow3-\left(\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\right)\ge3-\dfrac{ab+bc+ca}{2}\) (1)
Theo hệ quả của bất đẳng thức Cauchy
\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Rightarrow3\ge ab+bc+ca\)
\(\Rightarrow\dfrac{3}{2}\ge\dfrac{ab+bc+ca}{2}\)
\(\Rightarrow\dfrac{3}{2}\le3-\dfrac{ab+bc+ca}{2}\)(2)
Từ (1) và (2)
\(\Rightarrow3-\left(\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\right)\ge\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{a}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{c}{a^2+1}\ge\dfrac{3}{2}\) ( đpcm )
Dấu "=" xảy ra khi \(a=b=c=1\)
b) \(x,y\ge1\Rightarrow xy\ge1\)
BĐT đã cho tương đương với:
\(\left(\dfrac{1}{1+x^2}-\dfrac{1}{1+xy}\right)+\left(\dfrac{1}{1+y^2}-\dfrac{1}{1+xy}\right)\ge0\)
\(\Leftrightarrow\dfrac{xy-x^2}{\left(1+x^2\right)\left(1+xy\right)}+\dfrac{xy-y^2}{\left(1+y^2\right)\left(1+xy\right)}\ge0\)
\(\Leftrightarrow+\dfrac{x\left(y-x\right)}{\left(1+x^2\right)\left(1+xy\right)}+\dfrac{y\left(x-y\right)}{\left(1+y^2\right)\left(1+xy\right)}\ge0\)
\(\Leftrightarrow\dfrac{\left(y-x\right)^2\left(xy-1\right)}{\left(1+x^2\right)\left(1+y^2\right)\left(1+xy\right)}\ge0\)
BĐT cuối luôn đúng nên ta có đpcm
Đẳng thức xảy ra khi x=y hoặc xy=1
Ta có \(a\ge1;b\ge1\Rightarrow a\cdot b\ge1\) (1)
\(\Rightarrow\left(1+ab\right)\left(1+a^2\right)\left(1+b^2\right)>0\) (2)
Từ (1);(2)\(\Rightarrow\dfrac{\left(b-a\right)^2\left(ab-1\right)}{\left(1+ab\right)\left(1+a^2\right)\left(1+b^2\right)}\ge0\)
\(\Leftrightarrow\dfrac{b-a}{1+ab}\left(\dfrac{b^2\cdot a-a^2b-b+a}{\left(1+a^2\right)\left(1+b^2\right)}\right)\ge0\)
\(\Leftrightarrow\dfrac{b-a}{1+ab}\left(\dfrac{a}{1+a^2}-\dfrac{b}{1+b^2}\right)\ge0\)
\(\Leftrightarrow\dfrac{ab-a^2}{\left(1+ab\right)\left(1+a^2\right)}-\dfrac{b^2-ab}{\left(1+ab\right)\left(1+b^2\right)}\ge0\)
\(\Leftrightarrow\dfrac{ab-a^2+1-1}{\left(1+ab\right)\left(1+a^2\right)}-\dfrac{b^2-1-ab+1}{\left(1+ab\right)\left(1+b^2\right)}\ge0\)
\(\Leftrightarrow\dfrac{1}{1+a^2}-\dfrac{1}{1+ab}+\dfrac{1}{1+b^2}-\dfrac{1}{1+ab}\ge0\)
\(\Rightarrow\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}\ge\dfrac{2}{1+ab}\) (đpcm)
P/S: x thay = a , y thay = b nha