For $x,y,z∈ℝ^{+}$,without using Hölder's inequality prove that $\frac{x}{\sqrt{x^2+8yz}}+\frac{y}{\sqrt{y^2+8xz}}+\frac{z}{\sqrt{z^2+8xy}}\geq1$.

Question

For $x,y,z∈ℝ^{+}$, prove that $\frac{x}{\sqrt{x^2+8yz}}+\frac{y}{\sqrt{y^2+8xz}}+\frac{z}{\sqrt{z^2+8xy}}\geq1$.

In this question solution used Hölder's inequality, but I am looking a solution without using Hölder's inequality.

I first tried using AM-GM inequality but not results clearly.

Then I used Cauchy-Schwarz inequality which gave,

$${(\sum_{cyc}x²)(\sum_{cyc}\frac{1}{x²+8yz})\geq(\sum_{cyc}\frac{x}{\sqrt{x²+8yz}})}^2$$

But I couldn't solve it. Can anyone help me out?

Answer 1

Okay, just to get this question removed from the unanswered queue - we are looking for an exponent $a$ such that: $$\dfrac{x}{\sqrt{x^2+8yz}}\geq\dfrac{x^a}{x^a+y^a+z^a}\iff x^a+y^a+z^a\geq x^{a-1}\sqrt{x^2+8yz}.$$

But note that: $$(x^a+y^a+z^a)^2\geq (x^a+2(yz)^{\frac a2})^2=x^{2a} + 4x^a(yz)^{\frac a2}+4(yz)^a\geq x^{2a} + 8x^{\frac a2}(yz)^{\frac{3a}{4}}.$$ For $a = \frac 43$ then, this is just $$x^{\frac 83} + 8x^{\frac 23}yz = x^{\frac 13}\sqrt{x^2+8yz} = x^{a-1}\sqrt{x^2+8yz}$$ as desired.

Answer 2

By C-S twice and by AM-GM again we obtain: $$\sum_{cyc}\frac{x}{\sqrt{x^2+8yz}}=\sum_{cyc}\frac{x^2}{x\sqrt{x^2+8yz}}\geq\frac{(x+y+z)^2}{\sum\limits_{cyc}x\sqrt{x^2+8yz}}\geq$$ $$\geq\frac{(x+y+z)^2}{\sqrt{(x+y+z)\sum\limits_{cyc}x(x^2+8yz)}}=\sqrt{\frac{(x+y+z)^3}{\sum\limits_{cyc}(x^3+8xyz)}}\geq1.$$

Answer 3

Another way.

By AM-GM, C-S and AM-GM again we obtain:$$\sum_{cyc}\frac{x}{\sqrt{x^2+8yz}}=\sum_{cyc}\frac{2x(x+y+z)}{2\sqrt{(x+y+z)^2(x^2+8yz)}}\geq\sum_{cyc}\frac{2x(x+y+z)}{(x+y+z)^2+x^2+8yz}=$$ $$=\sum_{cyc}\frac{2x^2(x+y+z)}{x(x+y+z)^2+x^3+8xyz}\geq\frac{2(x+y+z)^3}{\sum\limits_{cyc}(x(x+y+z)^2+x^3+8xyz}=$$ $$=\frac{2(x+y+z)^3}{(x+y+z)^3+x^3+y^3+z^3+24xyz}\geq1.$$