Question

[Nguyễn Trần Khánh linh]

Help mình vs ạ:))

Answer 1

[Nguyễn Việt Lâm]

Áp dụng BĐT Bunhiacopxki (cho tất cả các bài):

1.

\(\left(3x+4y\right)^2\le\left(3^2+4^2\right)\left(x^2+y^2\right)=25\)

\(\Rightarrow\left|3x+4y\right|\le5\)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(\dfrac{3}{5};\dfrac{4}{5}\right)\)

2.

\(\left(x+2y\right)^2=\left(1.x+\sqrt{2}.\sqrt{2y}\right)^2\le\left(1+2\right)\left(x^2+2y^2\right)=3\)

\(\Rightarrow\left|x+2y\right|\le\sqrt{3}\)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(\dfrac{1}{\sqrt{3}};\dfrac{1}{\sqrt{3}}\right)\)

Answer 2

[Nguyễn Việt Lâm]

4.

a.

Áp dụng Bunhiacopxki:

\(\left(b+c+c+a+a+b\right)\left(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\right)\ge\left(a+b+c\right)^2\)

\(\Rightarrow2\left(a+b+c\right)\left(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\right)\ge\left(a+b+c\right)^2\)

\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{a+b+c}{2}\)

Dấu "=" xảy ra khi \(a=b=c\)