Question

[Nguyễn Ngọc Bảo Quang]

Cho x,y,z là số thực dương. CM: \(\Sigma\frac{xy^3z^3}{\left(x^2+yz\right)^2\left(y^3+z^3\right)}\le\frac{3}{8}\)

Answer 1

[Akai Haruma]

Lời giải:

Ta thấy:

$2(y^3+z^3)-(y^2+z^2)(y+z)=(y-z)^2(y+z)\geq 0$ với $y,z>0$

$\Rightarrow y^3+z^3\geq \frac{(y^2+z^2)(y+z)}{2}$

$\Rightarrow \frac{xy^3z^3}{(x^2+yz)^2(y^3+z^3)}\leq \frac{2xy^3z^3}{(x^2+yz)^2(y+z)(y^2+z^2)}$

Áp dụng BĐT AM-GM:

$(x^2+yz)(y+z)\geq 2x\sqrt{yz}.2\sqrt{yz}=4xyz$

$(x^2+yz)(y^2+z^2)=(x^2y^2+x^2z^2+yz^3+y^3z)\geq x^2y^2+x^2z^2+2y^2z^2$

$\Rightarrow \frac{2xy^3z^3}{(x^2+yz)^2(y+z)(y^2+z^2)}\leq \frac{1}{2}.\frac{y^2z^2}{(x^2y^2+y^2z^2)+(x^2z^2+y^2z^2)}$

$\leq \frac{1}{2}.\frac{1}{4}\left(\frac{y^2z^2}{x^2y^2+y^2z^2}+\frac{y^2z^2}{x^2z^2+y^2z^2}\right)$ (theo BĐT Cauchy-Schwarz)

$=\frac{1}{8}\left(\frac{y^2z^2}{x^2y^2+y^2z^2}+\frac{y^2z^2}{x^2z^2+y^2z^2}\right)$

Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế:

\(\Rightarrow \sum \frac{xy^3z^3}{(x^2+yz)^2(y^3+z^3)}\leq \frac{2xy^3z^3}{(x^2+yz)^2(y^2+z^2)(y+z)}\leq \frac{1}{8}\sum \left(\frac{y^2z^2}{x^2y^2+y^2z^2}+\frac{y^2z^2}{x^2z^2+y^2z^2}\right)=\frac{3}{8}\)

Ta có đpcm.

Dấu "=" xảy ra khi $x=y=z$

Answer 2

[tthnew]

Sau khi quy đồng ta cần chứng minh

$\sum \left( {\frac {43051\,{x}^{12}{y}^{5}{z}^{2}}{15384}}+{\frac {268775
\,{x}^{6}{y}^{8}{z}^{5}}{5128}}+{\frac {289287\,{x}^{6}{y}^{9}{z}^{4}
}{20512}}+{\frac {942859\,{x}^{7}{y}^{5}{z}^{7}}{41024}}+{\frac {
196483\,{x}^{7}{y}^{7}{z}^{5}}{3846}}+{\frac {21923\,{x}^{7}{y}^{9}{z}
^{3}}{41024}}+{\frac {16733\,{x}^{8}y{z}^{10}}{1282}}+{\frac {611699\,
{x}^{8}{y}^{6}{z}^{5}}{15384}}+{\frac {16733\,{x}^{8}{y}^{7}{z}^{4}}{
5128}}+{\frac {51295\,{x}^{8}{y}^{10}z}{20512}}+{\frac {2533405\,{x}^{
9}{y}^{5}{z}^{5}}{41024}}+{\frac {84531\,{x}^{9}{y}^{9}z}{10256}}+{
\frac {5305\,{x}^{10}y{z}^{8}}{20512}}+{\frac {451705\,{x}^{10}{y}^{4}
{z}^{5}}{10256}}+{\frac {43051\,{x}^{4}{y}^{2}{z}^{13}}{7692}}+{\frac
{268775\,{x}^{4}{y}^{5}{z}^{10}}{10256}}+{\frac {778763\,{x}^{4}{y}^{8
}{z}^{7}}{10256}}+17\,{x}^{5}{y}^{5}{z}^{9}+{\frac {2500385\,{x}^{5}{y
}^{7}{z}^{7}}{41024}}+{\frac {161613\,{x}^{6}y{z}^{12}}{20512}}+{
\frac {375259\,{x}^{6}{y}^{2}{z}^{11}}{15384}}+{\frac {942859\,{x}^{6}
{y}^{6}{z}^{7}}{20512}}+{\frac {54461\,x{y}^{5}{z}^{13}}{10256}}+{
\frac {8309\,x{y}^{8}{z}^{10}}{10256}}+{\frac {54461\,{x}^{2}{y}^{4}{z
}^{13}}{5128}}+{\frac {56585\,{x}^{2}{y}^{6}{z}^{11}}{10256}}+{\frac {
312143\,{x}^{2}{y}^{8}{z}^{9}}{30768}}+{\frac {16733\,{x}^{2}{y}^{9}{z
}^{8}}{2564}}+{\frac {3101\,{x}^{2}{y}^{12}{z}^{5}}{15384}}+{\frac {
69925\,{x}^{3}{y}^{3}{z}^{13}}{20512}}+5\,{x}^{3}{y}^{4}{z}^{12}+{
\frac {402457\,{x}^{3}{y}^{5}{z}^{11}}{41024}}+{\frac {1692943\,{x}^{3
}{y}^{7}{z}^{9}}{123072}}+{\frac {832741\,{x}^{3}{y}^{9}{z}^{7}}{20512
}}+{\frac {830539\,{x}^{3}{y}^{10}{z}^{6}}{61536}}+{\frac {398737\,{x}
^{10}{y}^{7}{z}^{2}}{30768}}+{\frac {84531\,{x}^{10}{y}^{8}z}{20512}}+
{\frac {7075\,{x}^{11}y{z}^{7}}{10256}}+{\frac {379913\,{x}^{11}{y}^{3
}{z}^{5}}{20512}}+{\frac {80021\,{x}^{11}{y}^{4}{z}^{4}}{30768}}+{
\frac {503849\,{x}^{11}{y}^{5}{z}^{3}}{61536}}+6\,{x}^{11}{y}^{7}z+3\,
{x}^{12}{y}^{2}{z}^{5}+3\,{y}^{10}{z}^{9}+{\frac {8309\,{x}^{10}{z}^{9
}}{5128}}+{\frac {7075\,{y}^{7}{z}^{12}}{5128}} \right) \left( x-y
\right) ^{2} \geq 0$