Với mọi số thực a,b,c. CMR: \(a^2+2b^2-2ab+2a-4b+2\ge0\)
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Với mọi số thực a,b,c. CMR: \(a^2+2b^2-2ab+2a-4b+2\ge0\)
\(VT=a^2+b^2+1-2ab+2a-2b+b^2-2b+1\)
\(VT=\left(a-b+1\right)^2+\left(b-1\right)^2\ge0\) (đpcm)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a=0\\b=1\end{matrix}\right.\)
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Chứng minh rằng với mọi số a,b,c ta luôn có :
a) a2 + 5b2 - 4ab + 2a - 6b + 3 > 0
b) a2 + 2b - 2ab + 2a - 4b + 2 >0
Cho a, b, c thỏa mãn: c\(\ne\)2b; a+b\(\ne\)\(\frac{c}2\); c2=4(ac+bc-2ab).
CMR: \(\frac{4a^2+(2a-c)^2}{4b^2+(2b-c)^2}=\frac{2a-c}{2b-c}\).
Tìm số nguyên a, b sao cho:
a)2ab+a+4b=5
b)6a-b+2ab=7
c)3a^2-3ab-2a=6-2b
d)2ab+a+b=2
Cho a,b,c t/m; c \(\ne\)2b, a + b \(\ne\) \(\frac{c}{2}\), c2 = 4(ac + bc - 2ab)
CMR: \(\frac{4a^2+\left(2a-c\right)^2}{4b^2+\left(2b-c\right)^2}=\frac{2a-c}{2b-c}\)
cho a, ,b ,c là số thực dương. CMR:
\(\frac{a^2-bc}{2a^2+b^2+c^2}+\frac{b^2-ca}{2b^2+c^2+a^2}+\frac{c^2-ab}{2c^2+a^2+b^2}\ge0\)
\(\frac{a^2-bc}{2a^2+b^2+c^2}+\frac{b^2-ca}{2b^2+c^2+a^2}+\frac{c^2-ab}{2c^2+a^2+b^2}\)
= \(\frac{1}{2}\left(\frac{2a^2-2bc}{2a^2+b^2+c^2}+\frac{2b^2-2ca}{2b^2+c^2+a^2}+\frac{2c^2-2ab}{2c^2+a^2+b^2}\right)\)
= \(\frac{1}{2}\left(\frac{2a^2-2bc}{2a^2+b^2+c^2}-1+\frac{2b^2-2ca}{2b^2+c^2+a^2}-1+\frac{2c^2-2ab}{2c^2+a^2+b^2}-1\right)+\frac{3}{2}\)
= \(-\frac{1}{2}\left(\frac{\left(b+c\right)^2}{2a^2+b^2+c^2}+\frac{\left(a+c\right)^2}{2b^2+c^2+a^2}+\frac{\left(a+b\right)^2}{2c^2+a^2+b^2}\right)+\frac{3}{2}\)
NHận xét:
\(\frac{\left(b+c\right)^2}{2a^2+b^2+c^2}\)\(=\frac{\left(b+c\right)^2}{\left(a^2+b^2\right)+\left(a^2+c^2\right)}\le\frac{b^2}{a^2+b^2}+\frac{c^2}{a^2+c^2}\)
Tương tự: \(\frac{\left(a+c\right)^2}{2b^2+c^2+a^2}\le\text{}\text{}\frac{a^2}{b^2+a^2}+\frac{c^2}{b^2+c^2}\)
\(\frac{\left(a+b\right)^2}{2c^2+a^2+b^2}\le\text{}\text{}\frac{a^2}{c^2+a^2}+\frac{b^2}{b^2+c^2}\)
=> \(\frac{\left(b+c\right)^2}{2a^2+b^2+c^2}+\frac{\left(a+c\right)^2}{2b^2+c^2+a^2}+\frac{\left(a+b\right)^2}{2c^2+a^2+b^2}\le3\)
=> \(-\frac{1}{2}\left(\frac{\left(b+c\right)^2}{2a^2+b^2+c^2}+\frac{\left(a+c\right)^2}{2b^2+c^2+a^2}+\frac{\left(a+b\right)^2}{2c^2+a^2+b^2}\right)+\frac{3}{2}\ge-\frac{1}{2}.3+\frac{3}{2}=0\)
=> \(\frac{a^2-bc}{2a^2+b^2+c^2}+\frac{b^2-ca}{2b^2+c^2+a^2}+\frac{c^2-ab}{2c^2+a^2+b^2}\ge0\)
Dấu "=" xảy ra <=> a = b = c
Cho 2a+2b khác c; 2b khác c; \(c^2=4\left(ac+bc-2ab\right).Cmr:\dfrac{4a^2+\left(2a-c\right)^2}{4b^2+\left(2b-c\right)^2}=\dfrac{2a-c}{2b-c}\)
CMR với mọi số thực a,b ta luôn có:
\(ab\left(a-2\right)\left(b+6\right)+13a^2+4b^2-26a+24b+16\ge0\)
cho a,b,c là các số thực dương thỏa mãn abc=1 CMR (4a-1)/((2b+1)^2)+(4b-1)/((2c+1)^2)+(4c-1)/((2a+1)^2)>1