Question

1/cho số a >0 tìm GTNN của P = 2a +\(\frac{4}{a}\)+\(\frac{16}{a+2}\)

2/ cho a,b,c là số thực ϵ [0;\(\frac{1}{4}\)) chứng minh:

\(\sqrt{a\left(1-4a\right)}+\sqrt{b\left(1-4b\right)}+\sqrt{c\left(1-4c\right)}\le\frac{3}{4}\)

3/ cho các số dương a,b,c tỏa abc = 1. Chứng minh

\(\frac{1}{a^2c+b^2c+1}+\frac{1}{b^2a+c^2a+1}+\frac{1}{c^2b+a^2b+1}\le1\)

Answer 1

1)

\(2a+\frac{4}{a}+\frac{16}{a+2}=\left(a+\frac{4}{a}\right)+\left[\left(a+2\right)+\frac{16}{a+2}\right]-2\ge4+8-2=10\)

Dấu "=" xảy ra khi a=2

Answer 2

2)

\(\hept{\begin{cases}\sqrt{a\left(1-4a\right)}=\frac{1}{2}\sqrt{4a\left(1-4a\right)}\le\frac{1}{2}\cdot\frac{4a+1-4a}{2}=\frac{1}{4}\\\sqrt{b\left(1-4b\right)}=\frac{1}{2}\sqrt{4\left(1-4a\right)}\le\frac{1}{2}\cdot\frac{4b+1-4b}{2}=\frac{1}{4}\\\sqrt{c\left(1-4c\right)}=\frac{1}{2}\sqrt{4c\left(1-4c\right)}\le\frac{1}{2}\cdot\frac{4c+1-4c}{2}=\frac{1}{4}\end{cases}}\)

\(\Rightarrow\sqrt{a\left(1-4a\right)}+\sqrt{b\left(1-4b\right)}+\sqrt{c\left(1-4c\right)}\le\frac{3}{4}\)

Dấu "=" xảy ra khi \(a=b=c=\frac{1}{8}\)

Answer 3

2)

Sửa lại:\(\sqrt{b\left(1-4b\right)}=\frac{1}{2}\sqrt{4b\left(1-4b\right)}\le\frac{1}{2}\cdot\frac{4b+1-4b}{2}=\frac{1}{4}\)

Mình đánh máy nhầm

Answer 4

3)

\(b^2c+ca^2+1=b^2c+ca^2+abc=c\left[\left(a-b\right)^2+3ab\right]\ge3abc\)

Tương tự ta cũng có: \(\hept{\begin{cases}c^2a+ab^2+1\ge3bc\\a^2b+bc^2+1\ge3abc\end{cases}}\)

\(\Rightarrow\Sigma_{cyc}\frac{1}{b^2c+ca^2+1}\le\frac{3}{3abc}=1\left(abc=1\right)\)

Dấu "=" xảy ra khi a=b=c=1