Cho x,y,z>0 và x+y+z=3. CMR: \(\frac{x^3}{y^3+8}+\frac{y^3}{z^3+8}+\frac{z^3}{x^3+8}\ge\frac{1}{9}+\frac{2}{27}\left(xy+yz+xz\right)\)
cho a+b+c=3
CMR \(\frac{a^3}{b^3+8}+\frac{b^3}{c^3+8}+\frac{c^3}{a^3+8}>=\frac{1}{3}\)
cho a,b,c>0. chứng minh: \(\frac{a^8+b^8+c^8}{a^3+b^3+c^3}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
cho a,b ,c deu duong .cmr
\(\frac{8}{\left(a+b\right)^2+4abc}+\frac{8}{\left(b+c\right)^2+4abc}+\frac{8}{\left(a+c\right)^2+4abc}+a^2+b^2+c^2\ge\frac{8}{a+3}+\frac{8}{b+3}+\frac{8}{c+3}\)
\(\sqrt[3]{\frac{1}{4}+\frac{\sqrt{5}}{8}}-\sqrt[3]{\frac{\sqrt{5}}{8}-\frac{1}{4}}=\)
cho số thực a,b,c>0. CMR
\(\frac{8}{\left(a+b\right)^2+4abc}+\frac{8}{\left(b+c\right)^2+4abc}+\frac{8}{\left(c+a\right)^2+4abc}+a^2+b^2+c^2\ge\frac{8}{a+3}+\frac{8}{b+3}+\frac{8}{c+3}\)
Tìm min,max của P=xyz biết A= \(\frac{8-x^2}{16+x^4}+\frac{8-y^2}{16+y^4}+\frac{8-z^2}{16+z^4}\ge0.\)
Cho a;b;c >0 thỏa mã \(a+b+c\le3\)Tìm min P \(=\left(3+\frac{1}{a}+\frac{1}{b}\right)\left(3+\frac{1}{b}+\frac{1}{c}\right)\left(3+\frac{1}{c}+\frac{1}{a}\right)\)
\(9+\frac{1}{8+\frac{2}{7+\frac{3}{6+\frac{4}{5+\frac{5}{4+\frac{6}{3+\frac{7}{2+\frac{8}{9}}}}}}}}\)
Biểu diễn dưới dạng phân số
Cho các số thực dương thỏa mãn:x+y+z=3
Tìm Min \(P=\left(\frac{x^3}{y^3+8}+\frac{y^3}{z^3+8}+\frac{z^3}{x^3+8}\right)-\frac{2\left(xy+yz+zx\right)}{27}\)