Proving inequality containing multiple square roots

Question

$$0.8(\sqrt{324-x}-\sqrt{289-x})>0.2(\sqrt{400-x}-\sqrt{361-x})$$

Does anyone know why this condition is satisfied for all x>0? This equation was given in my lecture slides and I can't wrap my head around it?

They derived this equation from it (in case it helps): $$0.8\times35 (\frac{\sqrt{324-x}-\sqrt{289-x}}{35})>0.2\times39 (\frac{\sqrt{400-x}-\sqrt{361-x}}{39})$$

Thank you in advance!

Answer 1

Let $289-x=t$.

Thus, $t\geq0$ and we need to prove that: $$4\left(\sqrt{35+t}-\sqrt{t}\right)>\sqrt{111+t}-\sqrt{72+t}$$ or $$\frac{4\cdot35}{\sqrt{35+t}+\sqrt{t}}>\frac{39}{\sqrt{111+t}+\sqrt{72+t}}$$ or $$140\left(\sqrt{111+t}+\sqrt{72+t}\right)>39\left(\sqrt{35+t}+\sqrt{t}\right),$$ which is obvious.

Answer 2

Let's define $f(x)=0.8(\sqrt{324-x}-\sqrt{289-x})$ and $g(x)=0.2(\sqrt{400-x}-\sqrt{361-x})$, $x\in (0,289]$. Now let's take derivative of both of these $$f'(x)=\frac{2}{5\sqrt{289-x}}-\frac{2}{5\sqrt{324-x}}>0$$ $$g'(x)=\frac{1}{10\sqrt{361-x}}-\frac{1}{10\sqrt{400-x}}>0$$ As you can see since denimonator is clearly larger in the fractions with negative sign in front, so we can conclude that both derivatives are larger than $0$(if you are unsure you can check for yourself it is a simple irrational inequality), which means that $f(x)$ and $g(x)$ are both monotonically increasing on $(0,289]$.

Now we need to show that $f'(x)>g'(x)$, or $\frac{2}{5\sqrt{289-x}}-\frac{2}{5\sqrt{324-x}}>\frac{1}{10\sqrt{361-x}}-\frac{1}{10\sqrt{400-x}}$. If we put that $t=289-x,x\in(0,289]$. We get that $$\frac{2}{5\sqrt{t}}-\frac{2}{5\sqrt{35-t}}>\frac{1}{10\sqrt{72-t}}-\frac{1}{10\sqrt{111-t}}\implies (10\sqrt{35-t}-10\sqrt{t})(100\sqrt{(72-t)(111-t)})<(10\sqrt{72-t}-10\sqrt{111-t})(25\sqrt{t(35-t)})$$ which is clearly true.

Now to finalize our proof we need to show that $\int_{0}^{289}{f'(x)}>\int_{0}^{289}{g'(x)}$. Which is true, since $\int_{0}^{289}{f'(x)}=\frac{4\sqrt{35}-4}{5}>\frac{\sqrt{111}-6\sqrt{2}-1}{5}=\int_{0}^{289}{g'(x)}$. Hence we can conclude that $f(x)>g(x),\forall x \in (0,289]$, meaning your inequality is true.

Proof was corrected with the help of Calvin Lin (check comments)

Keep in mind you could also prove this by solving the irrational inequality that's given and the result should give you the solution $\forall x \in [0,289]$, but that of course is tedious process.

Answer 3

We will show that $ f(x) = \frac{ \sqrt{ 324 - x } - \sqrt{ 289 - x } } { \sqrt{ 400 - x } - \sqrt{ 361 - x } } \geq 1$ on $x \in [ 0, 289]$, which is much stronger than your result that $f(x) > \frac{1}{4}$.

Observe that $f(0) = \frac{18 - 17} { 20 - 19 } = 1$.
It remains to show that $f'(x) > 0 $, from which the conclusion follows.

Rationalizing the denominator, $ f(x) = \frac{ (\sqrt{ 324 - x } - \sqrt{ 289 - x }) ( \sqrt{ 400 - x } + \sqrt{ 361 - x } ) } { 39} $.
Let $ a = 400 - x, b = 361 - x, c = 324 - x, d = 289 - x$, we have $ a > b > c > d$.
Applying the product rule and chain rule on derivatives, $$ \begin{array} {r c l } 39 f'(x) & = & ( -\frac{1}{ 2\sqrt{c}} + \frac{1}{2\sqrt{d}} ) ( \sqrt{a} + \sqrt{b} ) + ( \sqrt{c} - \sqrt{d} ) ( -\frac{1}{2\sqrt{a} } - \frac{1}{ 2\sqrt{b} } ) \\ & = &\frac{1}{2} (\sqrt{a} + \sqrt{b} )(\sqrt{c} - \sqrt{d} ) ( \frac{1}{ \sqrt{cd} }- \frac{1}{ \sqrt{ab} } ) \\ & > & 0 \end{array} $$.