chung minh bat dang thuc: 1/22+1/32+1/42+...+1/n2<1 (n>=2; n thuoc N)
chung minh bat dang thuc S=1/31+1/32+1/33+...+1/60.CMR3/4<S<4/5.tra loi nhanh ho minh nha
Cho tam giac ABC bat ky , chung minh bat dang thuc 1/(BC+CA)^2 + 1/(BC+AB)^2 >= 1/(BC^2+CA.AB)
chung minh bat dang thuc 2(a^4+1) + (b^2 +1)^2>=2(ab+1)^2
Chung minh bat dang thuc sau \(\dfrac{1}{\sqrt{n}}< 2\left(\sqrt{n}-\sqrt{n+1}\right)\) voi n>=2,n thuoc N
chung minh bat dang thuc
a/ x^2+xy+y^2+1>0
b/ x^2+5y^2-4xy>10y-4
a/ x2 + xy + y2 + 1
= [x2 + 2.x.\(\dfrac{y}{2}\) + (\(\dfrac{y}{2}\) )2 ] + \(\dfrac{3y^2}{4}\) + 1
= ( x + \(\dfrac{y}{2}\) )2 + \(\dfrac{3y^2}{4}\) + 1
Vì \(\left(x+\dfrac{y}{2}\right)^2\) \(\ge\) 0 với mọi x;y
và \(\dfrac{3y^2}{4}\ge0\) với mọi x;y
=> \(\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^2}{4}\ge0\) với mọi x;y
=> \(\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^2}{4}+1>0\)
cho a,b,c thuoc R chung minh rang a2+b2_> 2ab (1) áp dung va chung minh bat dang thuc sau
(a2+1)(b2+1)(c2+1) _> 8abc
\(a^2+b^2\ge2ab\)
\(\Rightarrow a^2-2ab+b^2\ge0\)
\(\Rightarrow\left(a-b\right)^2\ge0\) (luôn đúng)
Vậy \(a^2+b^2\ge2ab\)
Áp dụng vào ta được :
\(a^2+1\ge2a\)
\(b^2+1\ge2b\)
\(c^2+1\ge2c\)
\(\Rightarrow\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge2a.2b.2c=8abc\)(ĐPCM)
chung minh neu a,b,c=(0;1) thi co it nhat mot trong cac bat dang thuc sau day dung:a(1-b) <1/4 ; b(1-c)<1/4 ; c(1-a)<1/4
chung minh neu a,b,c=(0;1) thi co it nhat mot trong cac bat dang thuc sau day dung:a(1-b) <1/4 ; b(1-c)<1/4 ; c(1-a)<1/4
Chứng minh rằng: M = 1/22 + 1/32 + 1/42 + ... + 1/n2 < 1