记一道凑数狗题

发布于 2020-12-11  521 次阅读


Problem:已知正数\(a,b,c\)满足\(x^5+y^5+z^5=3\),求证:\(\dfrac{x^4}{y^3}+\dfrac{y^4}{z^3}+\dfrac{z^4}{x^3}\geq 3\)

证明:$$10\dfrac{x^4}{y^3}+3(x^{10}+x^5y^5+x^5y^5)\geq 19x^{\frac{100}{19}}$$

同理列出其他的两个不等式,相加得到

$$10(\dfrac{x^4}{y^3}+\dfrac{y^4}{z^3}+\dfrac{z^4}{x^3}) + 3(x^5+y^5+z^5)^2 \geq 19(x^{\frac{100}{19}}+y^{\frac{100}{19}}+z^{\frac{100}{19}})$$

根据幂平均不等式知$$(\dfrac{x^{\frac{100}{19}}+y^{\frac{100}{19}}+z^{\frac{100}{19}}}{3})^{\frac{19}{100}}\geq (\dfrac{x^5+y^5+z^5}{3})^{\frac{1}{5}}=1$$

整理即得$$10(\dfrac{x^4}{y^3}+\dfrac{y^4}{z^3}+\dfrac{z^4}{x^3})\geq 3\times 19-3\times 3^2=30$$

于是$$\dfrac{x^4}{y^3}+\dfrac{y^4}{z^3}+\dfrac{z^4}{x^3}\geq 3$$


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