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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 Svacxơ:
\(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{cd}+\dfrac{1}{da}\ge\dfrac{4}{ab+bc+cd+da}\)
Áp dụng BĐT Cô-si:
\(\dfrac{4}{ab+bc+cd+da}\ge\dfrac{4}{a^2+b^2+c^2+d^2}\)
Ta cần c/m: \(\dfrac{4}{a^2+b^2+c^2+d^2}\ge a^2+b^2+c^2+d^2\)
\(\Rightarrow\left(a^2+b^2+c^2+d^2\right)^2\ge4\)
Áp dụng BĐT Svacxơ: \(\left(\dfrac{a^2}{1}+\dfrac{b^2}{1}+\dfrac{c^2}{1}+\dfrac{d^2}{1}\right)^2\ge\dfrac{\left(a+b+c+d\right)^{2^2}}{16}\)
mà a+b+c+d=4 nên: \(\dfrac{\left(a+b+c+d\right)^4}{16}\ge\dfrac{64}{16}=4=VP\)
Vậy ta có đpcm.
1a)\(\dfrac{a^2+b^2}{2}\ge\dfrac{\left(a+b\right)^2}{4}\)
\(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\)(luôn đúng)
b)\(\dfrac{a^2+b^2+c^2}{3}\ge\dfrac{\left(a+b+c\right)^2}{9}\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)(luôn đúng)
2a)\(a^2+\dfrac{b^2}{4}\ge ab\)
\(\Leftrightarrow a^2-ab+\dfrac{b^2}{4}\ge0\)
\(\Leftrightarrow a^2-2\cdot\dfrac{1}{2}b\cdot a+\left(\dfrac{1}{2}b\right)^2\ge0\)
\(\Leftrightarrow\left(a-\dfrac{1}{2}b\right)^2\ge0\)(luôn đúng)
b)Đã cm
c)\(a^2+b^2+1\ge ab+a+b\)
\(\Leftrightarrow2a^2+2b^2+2\ge2ab+2a+2b\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\)(luôn đúng)
Dấu bằng xảy ra khi a=b=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 - Schwarz
\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Rightarrow3\ge ab+bc+ca\)
\(\Rightarrow\dfrac{3}{2}\ge\dfrac{ab+bc+ca}{2}\)
\(\Rightarrow\dfrac{3}{2}\le3-\dfrac{ab+bc+ca}{2}\) ( 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 \(a=b=c=1\)
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