\(1=\left(1.a+2.2b\right)^2\le\left(1^2+2^2\right)\left(a^2+4b^2\right)=5\left(a^2+4b^2\right)\)

\(\Rightarrow a^2+4b^2\ge\frac{1}{5}\)

Dấu "=" khi \(a=b=\frac{1}{5}\)

Ta có:

\(a^2+4b^2=a^2+\frac{16b^2}{4}\ge\frac{\left(a+4b\right)^2}{5}=\frac{1}{5}\)

Dấu \("="\) xảy ra khi: \(\left\{{}\begin{matrix}a=\frac{1}{2}\\b=\frac{1}{8}\end{matrix}\right.\)

Áp dụng BĐT Cauchy-Schwarz ta có:

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

\(\Rightarrow5\left(a^2+4b^2\right)\ge\left(a+4b\right)^2\)

\(\Rightarrow5\left(a^2+4b^2\right)\ge\left(a+4b\right)^2=1^2=1\)

\(\Rightarrow5\left(a^2+4b^2\right)\ge1\Rightarrow a^2+4b^2\ge\dfrac{1}{5}\)

Đẳng thức xảy ra khi \(a=b=\dfrac{1}{5}\)

Đặt \(T=a^2+4b^2\)(1)

Vì a+4b=1 => a=1-4b

Thế vào (1) ta được: \(T=\left(1-4b\right)^2+4b^2=20b^2-8b+1\)

<=> \(T=20\left(b^2-2\cdot\frac{1}{5}\cdot b+\frac{1}{25}\right)+\frac{1}{5}=20\left(b-\frac{1}{5}\right)^2+\frac{1}{5}\)

=> \(T\ge\frac{1}{5}\left(đpcm\right)\)

trả lời

anh ơi cái anyf dùng bất đẳng thức

(ax+by)^2<= (a^2+b^2)(x^2+y^2) cũng được nhỉ

cách này nhanh hơn đó ạ

hok tốt

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có: \(\dfrac{4a+3c}{4b+3d}=\dfrac{4bk+3dk}{4b+3d}=k\)

\(\dfrac{4a-3c}{4b-3d}=\dfrac{4bk-3dk}{4b-3d}=k\)

Do đó: \(\dfrac{4a+3c}{4b+3d}=\dfrac{4a-3c}{4b-3d}\)

tham khảo

4a + 5 < 4b+5

<=> 4a +5 - 5 < 4b+5 - 5

<=> 4a < 4b

<=> a < b