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Let $a,b,c \in \mathbb{R}_+^*$. Prove that $$\frac{a}{\sqrt{a^2+8bc}} + \frac{b}{\sqrt{b^2+8ca}}+ \frac{c}{\sqrt{c^2+8ab}} \geqslant 1.$$

I tried to follow the proposed solution for this which depended on Hölder's inequality, but I'm a bit confused about how they came up with the expression. How I remember Hölder's is that it states that $$\sum_{i=1}^n |x_iy_i| \leqslant (\sum_{i=1}^n|x_i|^p)^{1/p}(\sum_{i=1}^n|y_i|^q)^{1/q}$$

and we need the same Conjugate property as in Young's inequality $\frac{1}{p} + \frac{1}{q} =1.$

What they had was $$(\sum \frac{a}{\sqrt{a^2+8bc}})(\sum \frac{a}{\sqrt{a^2+8bc}})(\sum a(a^2+8bc)) \geqslant (a+b+c)^3.$$

From here it was quite straightforward, but any clarification on how we can get this result from Hölder's would be appreciated.

user26857
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    There's a Holder inequality with three functions. The continuous version is discussed at https://math.stackexchange.com/questions/56235/h%C3%B6lders-inequality-with-three-functions – Angina Seng Jul 13 '20 at 15:41
  • Rewrite your last inequality as $$\begin{align}\left(\sum_\text{cyc},\left(\frac{a^{\frac{2}{3}}}{(a^2+8bc)^{\frac{1}{3}}}\right)^{\frac{3}{2}}\right)^{\frac{2}{3}}&,\left(\sum_\text{cyc},\left({a^{\frac{1}{3}}}{(a^2+8bc)^{\frac{1}{3}}}\right)^{3}\right)^{\frac{1}{3}}\&\geq \sum_\text{cyc},\left(\frac{a^{\frac{2}{3}}}{(a^2+8bc)^{\frac{1}{3}}}\right),\left({a^{\frac{1}{3}}}{(a^2+8bc)^{\frac{1}{3}}}\right),.\end{align}$$ – Batominovski Jul 13 '20 at 15:43

2 Answers2

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The Holder's inequality for two sequences it's the following.

Let $a_1$, $a_2$,..., $a_n$, $b_1$, $b_2$,..., $b_n$, $\alpha$ and $\beta$ be positive numbers. Prove that: $$(a_1+a_2+...+a_n)^{\alpha}(b_1+b_2+...+b_n)^{\beta}\geq$$$$\geq\left(\left(a_1^{\alpha}b_1^{\beta}\right)^{\frac{1}{\alpha+\beta}}+\left(a_2^{\alpha}b_2^{\beta}\right)^{\frac{1}{\alpha+\beta}}+...+\left(a_n^{\alpha}b_n^{\beta}\right)^{\frac{1}{\alpha+\beta}}\right)^{\alpha+\beta}.$$

For positives $a$, $b$ and $c$ by Holder we obtain: $$\left(\sum_{cyc}\frac{a}{\sqrt{a^2+8bc}}\right)^2\sum_{cyc}a(a^2+8bc)\geq$$ $$\geq\left(\sum_{cyc}\sqrt[3]{\left(\frac{a}{\sqrt{a^2+8bc}}\right)^2a(a^2+8bc)}\right)^3=(a+b+c)^3.$$ In our case $n=3$, $\alpha=2$ and $\beta=1$.

Michael Rozenberg
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  • Could you elaborate on where is $\sum_{cyc}a(a^2+8bc)$ coming from? –  Jul 13 '20 at 17:17
  • @Daniel We need it for destruction of the radical, of the fraction and we need a factor $a$ for getting $(a+b+c)^3$. See please my writing of Holder. – Michael Rozenberg Jul 13 '20 at 17:53
  • I don't quite get it. Destruction of what radical? It seems that we have three. Also why do we even want $(a+b+c)^3$? –  Jul 13 '20 at 19:19
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    @Daniel We have a sum of three fractions. All these fractions have radicals in denominators. We want to delete these radicals because we can get something simple. We want to get $(a+b+c)^3$ because it a simplest expression, which we can get by using Holder and to delete fractions and radicals. By the way, if you know Hebrew, I am ready to explain all these in the Zoom. – Michael Rozenberg Jul 13 '20 at 19:27
  • Thanks, I'll try to get an understanding of this, and unfortunately, I don't know Hebrew. –  Jul 13 '20 at 19:56
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Here is my solution to the problem, I have used Jensen's inequality. Let us consider a function f(x) = $$\sqrt\frac{(k/x)^2}{(k/x)^2 +8x}$$. Since, f is a convex function, by Jensen's inequality we have, $$f(\frac{bc+ca+ab}{3}) ⩽ \frac{1}{3}[f(bc)+f(ca)+f(ab)]$$ $$⇒\frac{[f(bc)+f(ca)+f(ab)]}{3}≽ f(\frac{bc+ca+ab}{3})≽f((abc)^⅔)\space\space[By AM-GM inequality.]$$ $$⇒f(bc)+f(ca)+f(ab)≽3*\sqrt\frac{(abc)^⅔}{(abc)^⅔ +8(abc)^⅔}$$ $$⇒f(bc)+f(ca)+f(ab)≽3*\sqrt\frac{1}{1 + 8}=3*(\frac{1}{3})=1$$ $$⇒\frac{a}{\sqrt{a^2+8bc}}+\frac{b}{\sqrt{b^2+8ca}}+\frac{c}{\sqrt{c^2+8ab}}≽ 1$$ Hence proved. PS: The function is not convex $$\forall x\space\epsilon\space\mathbb R^+ .$$ $$\forall k\space\epsilon\space\mathbb R^+\space\exists\space a\space\epsilon\space\mathbb R^+$$ such that $$\forall x\space\epsilon\space (0,a)$$ f(x) is concave. However, $$\forall x\space\epsilon\space [a,∞)$$ f(x) is convex.

Teufel
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