MHB Can inequality be proven with positive real numbers and fractions?

[anemone]

For the positive real numbers $x,\,y$ and $z$ that satisfy $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=3$, prove that

$\dfrac{1}{\sqrt{x^3+1}}+\dfrac{1}{\sqrt{y^3+1}}+\dfrac{1}{\sqrt{z^3+1}}\le \dfrac{3}{\sqrt{2}}$.

 

[MarkFL]

My solution:

Spoiler

Using cyclic symmetry, we find the extremum must occur for:

$$(x,y,z)=(1,1,1)$$

If we define:

$$f(x,y,z)=\frac{1}{\sqrt{x^3+1}}+\frac{1}{\sqrt{y^3+1}}+\frac{1}{\sqrt{z^3+1}}$$

then we obtain:

$$f(1,1,1)=\frac{3}{\sqrt{2}}$$

To determine the nature of the extremum, we pick another point on the constraint:

$$(x,y,z)=\left(2,\frac{4}{5},\frac{4}{5}\right)$$

And we find:

$$f\left(2,\frac{4}{5},\frac{4}{5}\right)=\frac{21+10\sqrt{105}}{63}<\frac{3}{\sqrt{2}}$$

Thus, the extremum is a maximum, and we may state:

$$f(x,y,z)\le\frac{3}{\sqrt{2}}$$

 

[anemone]

My solution:

Spoiler

Using cyclic symmetry, we find the extremum must occur for:

$$(x,y,z)=(1,1,1)$$

If we define:

$$f(x,y,z)=\frac{1}{\sqrt{x^3+1}}+\frac{1}{\sqrt{y^3+1}}+\frac{1}{\sqrt{z^3+1}}$$

then we obtain:

$$f(1,1,1)=\frac{3}{\sqrt{2}}$$

To determine the nature of the extremum, we pick another point on the constraint:

$$(x,y,z)=\left(2,\frac{4}{5},\frac{4}{5}\right)$$

And we find:

$$f\left(2,\frac{4}{5},\frac{4}{5}\right)=\frac{21+10\sqrt{105}}{63}<\frac{3}{\sqrt{2}}$$

Thus, the extremum is a maximum, and we may state:

$$f(x,y,z)\le\frac{3}{\sqrt{2}}$$

Good job, MarkFL! And thanks for participating!

My solution:

Spoiler

By AM-GM, we have $x^3+1\ge 2x\sqrt{x}$ so $\dfrac{1}{\sqrt{x^3+1}}\le\dfrac{1}{\sqrt{2}\sqrt{x}\sqrt[4]{x}}$. By the same token we also have $\dfrac{1}{\sqrt{y^3+1}}\le\dfrac{1}{\sqrt{2}\sqrt{y}\sqrt[4]{y}}$ and $\dfrac{1}{\sqrt{z^3+1}}\le\dfrac{1}{\sqrt{2}\sqrt{z}\sqrt[4]{z}}$.

Adding the three inequalities we get:

$\dfrac{1}{\sqrt{x^3+1}}+\dfrac{1}{\sqrt{y^3+1}}+\dfrac{1}{\sqrt{z^3+1}}\le\dfrac{1}{\sqrt{2}}\left(\dfrac{1}{\sqrt{x}\sqrt[4]{x}}+\dfrac{1}{\sqrt{y}\sqrt[4]{y}}+\dfrac{1}{\sqrt{z}\sqrt[4]{z}}\right)$

Note that the following can be obtained by Cauchy–Schwarz inequality:

$\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\le\sqrt{1+1+1}\sqrt{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}=\sqrt{3}\sqrt{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}$

$\dfrac{1}{\sqrt{x}\sqrt[4]{x}}+\dfrac{1}{\sqrt{y}\sqrt[4]{y}}+\dfrac{1}{\sqrt{z}\sqrt[4]{z}}\le\sqrt{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}\sqrt{\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}}=\sqrt{3}\sqrt{\left(\sqrt{3}\sqrt{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}\right)}$

Putting these pieces together, and since we're told that $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=3$ we see that we have proved:

$\begin{align*}\dfrac{1}{\sqrt{x^3+1}}+\dfrac{1}{\sqrt{y^3+1}}+\dfrac{1}{\sqrt{z^3+1}}&\le\dfrac{1}{\sqrt{2}}\left(\dfrac{1}{\sqrt{x}\sqrt[4]{x}}+\dfrac{1}{\sqrt{y}\sqrt[4]{y}}+\dfrac{1}{\sqrt{z}\sqrt[4]{z}}\right)\\&\le \dfrac{1}{\sqrt{2}} \sqrt{3}\sqrt{\left(\sqrt{3}\sqrt{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}\right)}\\&\le \dfrac{3}{\sqrt{2}}\end{align*}$

 

[MarkFL]

Good job, MarkFL! And thanks for participating!...

Spoiler

You likely knew I would use cyclic symmetry just as surely as I knew you would use AM-GM. (Smirk)

 

[anemone]

Spoiler

You likely knew I would use cyclic symmetry just as surely as I knew you would use AM-GM. (Smirk)

Spoiler

LOL! That is very true!