How can I prove that $$\sqrt{3}-\sqrt{2}-0.32<0$$ This is what I did :
We have : $$\sqrt3\simeq1.732 \; ; \; \sqrt2\simeq1.414$$ $$\sqrt3<1.733 \; ; \; \sqrt2>1.413$$ $$\sqrt3<1.733 \; ; \; -\sqrt2<-1.413$$ Then we get : $$\sqrt3+(-\sqrt2)<1.733+(-1.413)$$ $$\sqrt3-\sqrt2<0.32$$ $$\sqrt3-\sqrt2-0.32<0$$ What you think for this solution, and can you please give me an other one without using an approximate value.
On
First we carefully use algebra to rewrite the inequality in an equivalent form with no radicals. We use squaring to do this, so it is important to only square when both sides of the inequality are positive: \begin{align*} & \sqrt{3} - \sqrt{2} - .32 < 0 \\ \iff & \sqrt{3} < \sqrt{2} + .32 \\ \iff & 3 < \bigl( \sqrt{2} + .32 \bigr)^2 \\ \iff & 3 < 2 + .64 \sqrt{2} + 0.1024 \\ \iff & \frac{.8176}{.64} < \sqrt{2} \\ \iff &\left( \frac{.8176}{.64} \right)^2 < 2 \end{align*} The rest is simple arithmetic estimates. I used a calculator to decide how many digits to keep, and then did the following by hand: $$\left( \frac{.8176}{.64} \right)^2 < \left( \frac{.88}{.64} \right)^2 = \bigl(\frac{11}{8}\bigr)^2 = \frac{121}{64} < \frac{128}{64} = 2 $$
On
This answer does not go beyond three digits numbers.
The inequality is equivalent to $${1\over\sqrt{2}+\sqrt{3}}<{8\over 25}$$ i.e. $$\sqrt{2}+\sqrt{3}>{25\over 8}$$ Raising to the second power gives an equivalent inequality $$5+2\sqrt{6}>{625\over 64}$$ or $$\sqrt{6}> {305\over 128}\quad(*)$$ We have $${305\over 128}<{308\over 128}={77\over 32}=2+{13\over 32}$$ Next $$\left(2+{13\over 32}\right )^2=4+{13\over 8}+{13^2\over 32^2}\\ <4+{13\over 8}+{1\over 4}=6 -{1\over 8}<6$$ which concludes the proof of $(*).$
On
And a tedious approach, which could nominally be performed by hand.
Multiplying both sides by $\sqrt{3}+\sqrt{2}$ shows that it is equivalent to showing $1 < {8 \over 25} (\sqrt{3}+\sqrt{2})$.
If we can generate a sequence such that $x_k \uparrow \sqrt{3}, y_k \uparrow \sqrt{2}$, then we can stop whenever $1 < {8 \over 25} (x_k+y_k)$.
Suppose $n \ge 1$. Since the function $f(x) = {1 \over x^2} - {1 \over n}$ is convex and strictly decreasing (on $x>0$) we see that the sequence of iterates generated by Newton's method starting at $x=1$ (any number $\le \sqrt{n}$ will do), then $x_k \uparrow \sqrt{n}$.
In particular, with $x_0 = y_0 = 1$ we compute the sequence $x_{k+1} = {3 \over 2} x_k - {x_k^3 \over 6}$, $y_{k+1} = {3 \over 2} y_k - {y_k^3 \over 4}$.
All computations can be performed exactly in rational arithmetic.
Grinding through the computations (well, Octave is doing the work) shows that $1 < {8 \over 25} (x_3+y_3)$.
On
This calculation keeps the numbers small. \begin{align*} \sqrt3-\sqrt2 < 0.32 &\iff 25\sqrt2+8 > 25\sqrt3 \\ & \iff 25^2\cdot2+400\sqrt2+64 > 25^2\cdot3 \\ & \iff 400\sqrt2 > 25^2-64 \\ & \iff \sqrt2 > \frac{561}{400} \\ & \iff \sqrt2 - \frac75 > \frac1{400}. \end{align*} The last inequality is true, because $$ \big(\sqrt2\big)^2-\left(\frac75\right)^2 = 2-\frac{49}{25} = \frac1{25}, $$ whence $$ \sqrt2-\frac75 = \frac1{25}\left(\sqrt2 + \frac75\right)^{-1}\!\! > \frac1{25}\left(\frac32+\frac32\right)^{-1}\!\! = \frac1{75} > \frac1{400}. $$
You can argue like this. Suppose you want to prove that $$\sqrt{3} - \sqrt{2} < 0.32.$$ Since both sides of the inequality are positive, you can square both sides to get $$5 - 2 \sqrt{6} < 0.32^2.$$ Rearranging yields $$5 - 0.32^2 < 2 \sqrt{6}.$$ Now, both sides are positive again, so you can square both sides to get $$(5 - 0.32^2)^2 < 24.$$ Now, simply by expanding the square, we see that the inequality is $$25 - 10*(0.32^2) + 0.32^4 < 24 \Rightarrow 1 < 10*(0.32^2) - 0.32^4.$$ Now we can compute $$10*(0.32^2) = 10.24$$ then, you can just argue that $0.32^4 < 0.24$ and that would allow you to finish the proof. Notice that all of the inequalities also go in the opposite direction, since we were only squaring both sides when we know that they are both positive.