Bài toán:
Cho $x,y,z>0$ thỏa mãn:$ \dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=2$. Chứng minh rằng
$\dfrac{1}{x^2(2x+3y+z)}+\dfrac{1}{y^2(2y+3z+x)}+$ $\dfrac{1}{z^2(2z+3x+y)}\le \dfrac{1}{3}$
Lời giải:
Áp dụng liên tiếp BĐT AM-GM, Cauchy Schwarz cho 3 số ta có:
$12=6\sum \dfrac{1}{x^3}=\sum \left [ \dfrac{2}{x^3}+\left ( \dfrac{1}{x^3}+\dfrac{1}{x^3}+\dfrac{1}{y^3} \right )+\dfrac{1}{3}\left ( \dfrac{1}{x^3}+\dfrac{1}{x^3}+\dfrac{1}{z^3} \right ) \right ]\ge \sum \left [ \dfrac{2}{x^3}+\dfrac{3}{x^2y}+\dfrac{3}{3x^2z} \right ]=\sum \left [ \dfrac{4}{2x^3}+\dfrac{9}{3x^2y}+\dfrac{1}{x^2z} \right] \ge \sum \left [ \dfrac{(2+3+1)^2}{2x^3+3x^2y+x^2z} \right ]=\sum \dfrac{36}{x^2(2x+3y+z)}$
$\Rightarrow \sum \dfrac{1}{x^2(2x+3y+z)}\le \dfrac{1}{3}$
Dấu "=" xảy ra khi: $x=y=z=\sqrt[3][\dfrac{3}{2}]$
Cho $x,y,z>0$ thỏa mãn:$ \dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=2$. Chứng minh rằng
$\dfrac{1}{x^2(2x+3y+z)}+\dfrac{1}{y^2(2y+3z+x)}+$ $\dfrac{1}{z^2(2z+3x+y)}\le \dfrac{1}{3}$
Lời giải:
Áp dụng liên tiếp BĐT AM-GM, Cauchy Schwarz cho 3 số ta có:
$12=6\sum \dfrac{1}{x^3}=\sum \left [ \dfrac{2}{x^3}+\left ( \dfrac{1}{x^3}+\dfrac{1}{x^3}+\dfrac{1}{y^3} \right )+\dfrac{1}{3}\left ( \dfrac{1}{x^3}+\dfrac{1}{x^3}+\dfrac{1}{z^3} \right ) \right ]\ge \sum \left [ \dfrac{2}{x^3}+\dfrac{3}{x^2y}+\dfrac{3}{3x^2z} \right ]=\sum \left [ \dfrac{4}{2x^3}+\dfrac{9}{3x^2y}+\dfrac{1}{x^2z} \right] \ge \sum \left [ \dfrac{(2+3+1)^2}{2x^3+3x^2y+x^2z} \right ]=\sum \dfrac{36}{x^2(2x+3y+z)}$
$\Rightarrow \sum \dfrac{1}{x^2(2x+3y+z)}\le \dfrac{1}{3}$
Dấu "=" xảy ra khi: $x=y=z=\sqrt[3][\dfrac{3}{2}]$
Post a Comment