Chứng minh với x, y, z dương ta có x^3 / yz + y^3 / xz + x^3 / xy > = x + y + z

Ta có:

\(\frac{{{x^3}}}{{yz}} + \frac{{{y^3}}}{{xz}} + \frac{{{z^3}}}{{xy}} = \frac{{{x^4}}}{{xyz}} + \frac{{{y^4}}}{{xyz}} + \frac{{{z^4}}}{{xyz}} = \frac{{{x^4} + {y^4} + {z^4}}}{{xyz}}\)

Áp dụng bất đẳng thức \({a^2} + {b^2} + {c^2} \ge \frac{{{{\left( {a + b + c} \right)}^2}}}{3}\) ta có

\[{{\rm{x}}^4} + {y^4} + {z^4} \ge \frac{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^2}}}{3}\]

Suy ra \(\frac{{{x^4} + {y^4} + {z^4}}}{{xyz}} \ge \frac{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^2}}}{{3xyz}} \ge \frac{{{{\left[ {\frac{{{{\left( {x + y + z} \right)}^2}}}{3}} \right]}^2}}}{{\frac{{{{\left( {x + y + z} \right)}^3}}}{{3.3}}}}\)

\(\frac{{{x^4} + {y^4} + {z^4}}}{{xyz}} \ge \frac{{{{\left( {x + y + z} \right)}^4}}}{{{{\left( {x + y + z} \right)}^3}}} = x + y + z\)

Do đó \(\frac{{{x^3}}}{{yz}} + \frac{{{y^3}}}{{xz}} + \frac{{{z^3}}}{{xy}} \ge x + y + z\)

Vậy \(\frac{{{x^3}}}{{yz}} + \frac{{{y^3}}}{{xz}} + \frac{{{z^3}}}{{xy}} \ge x + y + z\).