# another proof of Young inequality

Let

$$F(x)={\int}_{0}^{x}\varphi (t)\mathit{d}t,\text{and}G(x)={\int}_{0}^{x}{\varphi}^{-1}(t)\mathit{d}t.$$ |

Since ${\varphi}^{-1}$ is strictly increasing^{}, $G$ is strictly convex, hence lies above its supporting line, i.e. for every $c$ and $x\ne c$

$$G(b)>G(c)+{G}^{\prime}(c)(b-c)=G(c)+{\varphi}^{-1}(c)(b-c).$$ |

In particular, for $c=\varphi (a)$ we have

$$F(a)+G(b)>F(a)+G(\varphi (a))+a(b-\varphi (a))=ab,$$ |

because $F(a)+G(\varphi (a))=a\varphi (a)$.

Title | another proof of Young inequality |
---|---|

Canonical name | AnotherProofOfYoungInequality |

Date of creation | 2013-03-22 15:45:38 |

Last modified on | 2013-03-22 15:45:38 |

Owner | a4karo (12322) |

Last modified by | a4karo (12322) |

Numerical id | 10 |

Author | a4karo (12322) |

Entry type | Proof |

Classification | msc 26D15 |