Question

Answer 1

\(\dfrac{1}{3x^2+y^2}+\dfrac{2}{y^2+3xy}=\dfrac{1}{3x^2+y^2}+\dfrac{4}{2y^2+6xy}\)

Áp dụng BĐT Svácxơ, ta có:

\(\dfrac{1}{3x^2+y^2}+\dfrac{4}{2y^2+6xy}\ge\dfrac{\left(1+2\right)^2}{3x^2+6xy+3y^2}=\dfrac{3}{\left(x+y\right)^2}\ge\dfrac{3}{1}=3\left(x+y\le1\right)\)

Answer 2

\(\dfrac{a}{b+c}+\dfrac{b}{c+d}+\dfrac{c}{d+a}+\dfrac{d}{a+b}=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{cb+db}+\dfrac{c^2}{dc+ac}+\dfrac{d^2}{ad+bd}\left(1\right)\)-Áp dụng BĐT Schwarz ta có:

\(\dfrac{a^2}{ab+ac}+\dfrac{b^2}{cb+db}+\dfrac{c^2}{dc+ac}+\dfrac{d^2}{ad+bd}\ge\dfrac{\left(a+b+c+d\right)^2}{ab+ac+cb+db+dc+ac+ad+bd}=\dfrac{\left(a+b+c+d\right)^2}{a\left(b+c\right)+d\left(b+c\right)+b\left(c+d\right)+a\left(c+d\right)}=\dfrac{\left(a+b+c+d\right)^2}{\left(b+c\right)\left(a+d\right)+\left(c+d\right)\left(a+b\right)}\left(2\right)\)-Áp dụng BĐT Caushy ta có:​\(\dfrac{\left(a+b+c+d\right)^2}{\left(b+c\right)\left(a+d\right)+\left(c+d\right)\left(a+b\right)}\ge\dfrac{\left(a+b+c+d\right)^2}{\dfrac{\left[\left(b+c\right)+\left(a+d\right)\right]^2}{4}+\dfrac{\left[\left(c+d\right)+\left(a+b\right)\right]^2}{4}}=\dfrac{\left(a+b+c+d\right)^2}{\dfrac{2\left(a+b+c+d\right)^2}{4}}=2\left(3\right)\)-Từ (1), (2), (3) ta suy ra đpcm.

-Dấu "=" xảy ra \(\Leftrightarrow a=b=c=d\)​