# Generalization of Hölder’s inequality with negative exponent

How can I prove that (1a+1b+1c)(√a+√b+√c)2≥33(\frac1a+\frac1b+\frac1c)(\sqrt a+\sqrt b+\sqrt c)^2\ge 3^3 using Hأ¶lder’s Inequality

This cannot be done with the usual Hأ¶lder inequality, there is a negative exponent and 22 between 00 and 11. And I couldn’t find any generalization with those exponents. (There’s one formula which holds for a different case here)

=================

1

The inequality itself can be proved by AM-GM:1a+1b+1c≥3(abc)1/3\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq3(abc)^{1/3} and √a+√b+√c≥3(abc)1/6\sqrt{a}+\sqrt{b}+\sqrt{c}\geq3(abc)^{1/6}.
– yurnero
2 days ago

@yumero thanks but I want to apply Holder
– user257
2 days ago

Yes I know, hence that was a comment only.
– yurnero
2 days ago

=================

1

=================

Hأ¶lder is the following.

(In the following form much easier to use Hأ¶lder!)

Let ai>0a_i>0, bi>0b_i>0, α>0\alpha>0 and β>0\beta>0. Hence, we have:

(a1+a2+…+an)α(b1+b2+…+bn)β≥(a_1+a_2+…+a_n)^{\alpha}(b_1+b_2+…+b_n)^{\beta}\geq
≥((aα1bβ1)1α+β+(aα2bβ2)1α+β+…+(aαnbβn)1α+β)α+β\geq\left(\left(a_1^{\alpha}b_1^{\beta}\right)^{\frac{1}{\alpha+\beta}}+\left(a_2^{\alpha}b_2^{\beta}\right)^{\frac{1}{\alpha+\beta}}+…+\left(a_n^{\alpha}b_n^{\beta}\right)^{\frac{1}{\alpha+\beta}}\right)^{\alpha+\beta}

For your inequality n=3n=3, α=1\alpha=1 and β=2\beta=2 and it’s just Hأ¶lder!