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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 \(a.b\ge1\Rightarrowđpcm\)
Câu 1:
Ta có: \(\left(\dfrac{a+b}{2}\right)^2\ge ab\)
\(\Leftrightarrow\dfrac{\left(a+b\right)^2}{2^2}-ab\ge0\)
\(\Leftrightarrow\dfrac{a^2+2ab+b^2-4ab}{4}\ge0\)
\(\Leftrightarrow\dfrac{a^2-2ab+b^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\)
Vì \(\left(a-b\right)^2\ge0\forall a,b\)
\(\Rightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\forall a,b\)
\(\Rightarrow\left(\dfrac{a+b}{2}\right)^2\ge ab\) (1)
Ta có: \(\dfrac{a^2+b^2}{2}\ge\left(\dfrac{a+b}{2}\right)^2\)
\(\Leftrightarrow\dfrac{a^2+b^2}{2}-\dfrac{\left(a+b\right)^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{2a^2-2b^2-a^2-2ab-b^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{a^2-2ab-b^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\)
Vì \(\left(a-b\right)^2\ge0\forall a,b\)
\(\Rightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\forall a,b\)
\(\Rightarrow\dfrac{a^2+b^2}{2}\ge\left(\dfrac{a+b}{2}\right)^2\) (2)
Từ (1) và (2) \(\Rightarrow ab\le\left(\dfrac{a+b}{2}\right)^2\le\dfrac{a^2+b^2}{2}\)
5 , a3+b3+c3\(\ge\) 3abc
\(\Leftrightarrow\) a3+3a2b+3ab2+b3+c3-3a2b-3ab2-3abc\(\ge\) 0
\(\Leftrightarrow\) (a+b)3+c3-3ab(a+b+c) \(\ge0\)
\(\Leftrightarrow\) (a+b+c)(a2+2ab+b2-ac-bc+c2)-3ab(a+b+c) \(\ge0\)
\(\Leftrightarrow\) (a+b+c)(a2+b2+c2-ab-bc-ca)\(\ge0\) (1)
ta co : a,b,c>0 \(\Rightarrow\)a+b+c>0 (2)
(a-b)2+(b-c)2+(c-a)2\(\ge0\)
<=> 2a2+2b2+2c2-2ac-2cb-2ab\(\ge0\)
<=>a2+b2+c2-ab-bc-ac\(\ge\) 0 (3)
Từ (1)(2)(3)=> pt luôn đúng
a) Xét hiệu : VT - VP
= \(\dfrac{\left(a+b\right)^2}{4}\) _ ab = \(\dfrac{a^2+2ab+b^2}{4}\)- \(\dfrac{4ab}{4}\)
= \(\dfrac{a^2-2ab+b^2}{4}\) = \(\dfrac{\left(a-b\right)^2}{4}\)
Có : (a - b )2 \(\ge\) 0 => \(\dfrac{\left(a-b\right)^2}{4}\) \(\ge\) 0 .
(bất phương trình đúng ) .
=> VT - VP \(\ge\) 0 => ( \(\dfrac{a+b}{2}\))2 \(\ge\) ab .
b) Xét hiệu ; VP - VT
= \(\dfrac{a^2+b^2}{2}\)-(\(\dfrac{a+b}{2}\))2
= \(\dfrac{2a^2+2b^2-\left(a^2+2ab+b^2\right)}{4}\)
= \(\dfrac{\left(a-b\right)^2}{4}\) .
Có : (a-b)2 \(\ge\) 0 => \(\dfrac{\left(a-b\right)^2}{4}\) \(\ge\) 0 .
VP - VT \(\ge\) 0 .
Vậy ( \(\dfrac{a+b}{2}\) )2 \(\le\) \(\dfrac{a^2+b^2}{2}\) .
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 \(ab=1\)
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)
Ta có: \(x\ge1;y\ge1\)
\(\Rightarrow\left\{{}\begin{matrix}x\left(y-1\right)\ge0\\y\left(x-1\right)\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}xy\ge x^2\\xy\ge y^2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}1+xy\ge1+x^2\\1+xy\ge1+y^2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{1+x^2}\ge\dfrac{1}{1+xy}\left(1\right)\\\dfrac{1}{1+y^2}\ge\dfrac{1}{1+xy}\left(2\right)\end{matrix}\right.\)
Cộng vế với vế (1) và (2) ta được : {cái bđt ở đầu bài chép xuống đây}
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)
Bài 3:
a) Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{xy}+\frac{2}{x^2+y^2}=2\left(\frac{1}{2xy}+\frac{1}{x^2+y^2}\right)\) \(\geq 2.\frac{(1+1)^2}{2xy+x^2+y^2}=\frac{8}{(x+y)^2}=8\)
Dấu bằng xảy ra khi \(x=y=\frac{1}{2}\)
b) Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{xy}+\frac{1}{x^2+y^2}=\frac{1}{2xy}+\left (\frac{1}{2xy}+\frac{1}{x^2+y^2}\right)\geq \frac{1}{2xy}+\frac{(1+1)^2}{2xy+x^2+y^2}\)
\(=\frac{1}{2xy}+\frac{4}{(x+y)^2}\)
Theo BĐT AM-GM:
\(xy\leq \frac{(x+y)^2}{4}=\frac{1}{4}\Rightarrow \frac{1}{2xy}\geq 2\)
Do đó \(\frac{1}{xy}+\frac{1}{x^2+y^2}\geq 2+4=6\)
Dấu bằng xảy ra khi \(x=y=\frac{1}{2}\)
Bài 1: Thiếu đề.
Bài 2: Sai đề, thử với \(x=\frac{1}{6}\)
Bài 4 a) Sai đề với \(x<0\)
b) Áp dụng BĐT AM-GM:
\(x^4-x+\frac{1}{2}=\left (x^4+\frac{1}{4}\right)-x+\frac{1}{4}\geq x^2-x+\frac{1}{4}=(x-\frac{1}{2})^2\geq 0\)
Dấu bằng xảy ra khi \(\left\{\begin{matrix} x^4=\frac{1}{4}\\ x=\frac{1}{2}\end{matrix}\right.\) (vô lý)
Do đó dấu bằng không xảy ra , nên \(x^4-x+\frac{1}{2}>0\)
Bài 6: Áp dụng BĐT AM-GM cho $6$ số:
\(a^2+b^2+c^2+d^2+ab+cd\geq 6\sqrt[6]{a^3b^3c^3d^3}=6\)
Do đó ta có đpcm
Dấu bằng xảy ra khi \(a=b=c=d=1\)