Áp dụng \(x^2+y^2\ge2xy\)
Ta có: \(\left(a^2+\frac{1}{4}\right)+\left(b^2+\frac{1}{4}\right)\ge a+b\)
\(\Leftrightarrow a^2+b^2\ge a+b-\frac{1}{4}-\frac{1}{4}\)
\(\Leftrightarrow a^2+b^2\ge a+b-\frac{1}{2}\)
Dấu "=" xảy ra khi \(a=b=\frac{1}{2}\)
CM BĐT
a^2+b^2+2=2(a+b)
CM BĐT: \(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}\ge\dfrac{2}{ab+1}\) với \(ab\ge1\)
CM BĐT sau : \(\left(a^2+b^2\right)\left(a^2+1\right)\ge4a^2b\forall a,b\)
cm bđt a2/b2 + b2/a2 >= 2
Cm BĐT a2+b2+c2>=ab+2(a+b)
CM BĐT (a+b)(a^3+b^3)=< 2(a^4+b^4)
CM BĐT: \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\) với \(a,b,c>0\)
1/ Cm BĐT : a2 + 2b2 + c2 >= 2ab - 2ac + bc
2/ Cho a3 + b3 = a - b, cm a2 + ab +b2 <1
3/ Giải PT : (x2 - 20162)2 - 8064x - 1 = 0
Cho 3 số nguyên dương a , b, c thỏa mãn : \(a^2+b^2+c^2=\dfrac{5}{3}\)
CM BĐT : \(\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{c}< \dfrac{1}{abc}\)
