Question

I : CMR

a) (ab+1)^2 \(\ge\)\(\frac{4}{ab}\)

b) \(\left(ab+2\right)^2\le\left(a^2+1\right)\left(b^2+4\right)\)

c) \(2\left(4a^2+b^2\right)\ge\left(2a+b\right)^2\)

d) \(x^5+y^5\ge xy\left(x^3+y^3\right)\) với x , y >0

help me !!!

Answer 1

a ) Ta có : \(\left(ab+1\right)^2\ge4ab\)

\(\Leftrightarrow a^2b^2+2ab+1-4ab\ge0\)

\(\Leftrightarrow\left(ab-1\right)^2\ge0\)

=> BĐT luôn đúng

Dấu " = " xảy ra \(\Leftrightarrow ab=1\)

b ) Áp dụng BĐT Bunhiacopxki , ta có :

\(\left(ab+1.2\right)^2\le\left(a^2+1^2\right)\left(b^2+2^2\right)=\left(a^2+1\right)\left(b^2+4\right)\)

Dấu " = " xảy ra \(\Leftrightarrow2a=b\)

c ) Áp dụng BĐT Cô - si cho 2 số không âm , ta có :

\(4a^2+b^2\ge2\sqrt{4a^2.b^2}=4ab\)

\(\Rightarrow2\left(4a^2+b^2\right)\ge4a^2+4ab+b^2=\left(2a+b\right)^2\)

Dấu " = " xảy ra \(\Leftrightarrow2a=b\)

d ) \(x^5+y^5\ge xy\left(x^3+y^3\right)\)

\(\Leftrightarrow x^5-x^4y-y^4x+y^5\ge0\)

\(\Leftrightarrow\left(x^4-y^4\right)\left(x-y\right)\ge0\)

\(\Leftrightarrow\left(x-y\right)^2\left(x+y\right)\left(x^2+y^2\right)\ge0\)

Vì x ; y > 0 => BĐT luôn đúng

Dấu " = " xảy ra \(\Leftrightarrow x=y\)