Same as $6^{\frac1{\sqrt 2}}4^{\frac1{\sqrt 3}}>5$ $6^{\frac1{\sqrt2}}>6^{\frac23}>3$ because $6^2>3^3$ $4^{\frac1{\sqrt3}}>4^{\frac12}=2$ $2\cdot3>5$ stupid

ythomashu wrote: Same as $6^{\frac1{\sqrt 2}}4^{\frac1{\sqrt 3}}>5$ No

pco wrote: ythomashu wrote: Same as $6^{\frac1{\sqrt 2}}4^{\frac1{\sqrt 3}}>5$ No ???

This is in fact equivalent to $6^{\sqrt 3}4^{\sqrt 2}>5^{{\sqrt 2}+\sqrt 3}$ (and not, as you likely did : $5^{\sqrt 2\sqrt 3}$ And so $6^{\frac 1{\sqrt 2}}4^{\frac 1{\sqrt 3}}>5^{\frac 1{\sqrt 2}+\frac 1{\sqrt 3}}$ Which is quite different

Lamp909 wrote: Prove that $$(\frac{6}{5})^{\sqrt{3}}>(\frac{5}{4})^{\sqrt{2}}$$. For $x\in(0,1)$ : $x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\frac{x^5}5$ $>\ln(1+x)>$ $x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4$ Setting $x=\frac 15$, we get $\ln \frac 65>\frac{1367}{7500}$ Setting $x=\frac 14$, we get $\ln\frac 54 < \frac{857}{3840}$ It then would be enough to check $\sqrt 3\frac{1367}{7500}>\sqrt 2\frac{857}{3840}$ Which is $\sqrt 3\times 1367\times 3840>\sqrt 2\times 857\times 7500$ And, simplifying : $\sqrt 3\times 1367\times 64>\sqrt 2\times 857\times 125$ And, squaring : $22962450432 > 22951531250$ Which is true. Q.E.D.

EDIT: Oh, sorry, my bad. I've messed up the first derivative, maybe because thinking of the second one at the same time... I don't know. EDIT: Nevertheless, the idea can be saved, see #15. That's, we must prove: $$6^{\sqrt{3}}4^{\sqrt{2}}>5^{\sqrt{3}+\sqrt{2}} $$ Consider the function $f(x):=(5+x)^{\sqrt{3}}(5-x)^{\sqrt{2}}$. It's enough to prove $f(x)$ is increasing when $x \in [0,1]$. Let's calculate $f'(x)$: $$f'(x)=(\sqrt{3}-1)(5+x)^{\sqrt{3}-1}(5-x)^{\sqrt{2}}-(\sqrt{2}-1)(5+x)^{\sqrt{3}}(5-x)^{\sqrt{2}-1}$$$$f'(x)=\left(5(\sqrt{3}-\sqrt{2})-(\sqrt{3}+\sqrt{2}-2)x \right)(5+x)^{\sqrt{3}-1}(5-x)^{\sqrt{2}-1}$$Now, it is easy to see $f'(x)>0\,,\,x\in(0,1)$, that's $f(1)>f(0)$ and the result follows.

dgrozev wrote: ... $$f'(x)=(\sqrt{3}-1)(5+x)^{\sqrt{3}-1}(5-x)^{\sqrt{2}}-(\sqrt{2}-1)(5+x)^{\sqrt{3}}(5-x)^{\sqrt{2}-1}$$.... Unfortunately no. Correct form is : : $f'(x)=\sqrt{3}(5+x)^{\sqrt{3}-1}(5-x)^{\sqrt{2}}-\sqrt{2}(5+x)^{\sqrt{3}}(5-x)^{\sqrt{2}-1}$ And this changes the conclusion $(f(x)$ is increasing over $(0,\sim 0.50510)$ and decreasing over $(\sim 0.50510,1)$)

pco wrote: Lamp909 wrote: Prove that $$(\frac{6}{5})^{\sqrt{3}}>(\frac{5}{4})^{\sqrt{2}}$$. For $x\in(0,1)$ : $x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\frac{x^5}5$ $>\ln(1+x)>$ $x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4$ Setting $x=\frac 15$, we get $\ln \frac 65>\frac{1367}{7500}$ Setting $x=\frac 14$, we get $\ln\frac 54 < \frac{857}{3840}$ It then would be enough to check $\sqrt 3\frac{1367}{7500}>\sqrt 2\frac{857}{3840}$ Which is $\sqrt 3\times 1367\times 3840>\sqrt 2\times 857\times 7500$ And, simplifying : $\sqrt 3\times 1367\times 64>\sqrt 2\times 857\times 125$ And, squaring : $22962450432 > 22951531250$ Which is true. Q.E.D. Very lengthy method

Smita wrote: Very lengthy method I do agree. Dont hesitate to post any smarter / shorter method : this is one interest of AOPS (providing many different solutions to a same problem)

I don't have a smarter method, but I do have some tricks to make the calculations less painful. Instead of the usual series for $\ln(1+x)$, we will use the series \[\ln\left(\frac{1+x}{1-x}\right)=2\left(x+\frac{x^3}3+\frac{x^5}5+\cdots\right),\]which converges much faster. Substituting $x=\frac19$ and $x=\frac1{11}$ respectively, and bounding the terms with some geometric progressions, we get: \begin{align*} \ln\frac54&<2\left(\frac19+\frac1{3\cdot9}\left(\frac1{9^2}+\frac1{9^4}+\cdots\right)\right)\\ &=2\left(\frac19+\frac1{3\cdot9\cdot80}\right)=\frac29+\frac1{1080},\\ \ln\frac65&>2\left(\frac1{11}+\frac1{3\cdot11^2}+\frac1{3^2\cdot11^5}+\cdots\right)\\ &=\frac2{11}\cdot\frac{363}{362}=\frac2{11}+\frac1{1991}. \end{align*}Then the desired inequality reduces to

chronondecay wrote: ... $\ln\frac65>2\left(\frac1{11}+\frac1{3\cdot11^2}+\frac1{3^2\cdot11^5}+\cdots\right)$... Little typo : $\ln\frac65>2\left(\frac1{11}+\frac1{3\cdot11^3}+\frac1{3^2\cdot11^5}+\cdots\right)$ But the next line is OK and so the result quite OK ... ... and smarter than mine, indeed. Congrats.

We are going to prove the general inequality:$$(\frac{2a+2}{2a+1})^{\sqrt{a+1}}>(\frac{2a+1}{2a})^{\sqrt{a}}$$for every $a\ge0$ , that gives the desired inequality for $a=2$. We will prove $\sqrt{a+1}(ln(2a+2)-ln(2a+1))>\sqrt{a}(ln(2a+1)-ln(2a))$ The key step of the solution is extracting the following equallity: $$D = \int^{a+1}_{a} \int^{a+1}_{a} {\frac{y-x}{2\sqrt{x}(x+y)^2}}\,dy\,dx = \sqrt{a+1}(ln(2a+2)-ln(2a+1))-\sqrt{a}(ln(2a+1)-ln(2a))$$Proof: $\frac{y-x}{2\sqrt{x}(x+y)^2}=\frac{1}{2\sqrt{x}(x+y)} - \frac{\sqrt{x}}{(x+y)^2}$ so, $ \int^{a+1}_{a} {\frac{y-x}{2\sqrt{x}(x+y)^2}}\,dy =\frac{ln(x+a+1)-ln(x+a)}{2\sqrt{x}} + \frac{\sqrt{x}}{x+a+1} - \frac{\sqrt{x}}{x+a}$ and since $\frac{d}{dx} [ \sqrt{x}(ln(x+a+1)-ln(x+a))] = \frac{ln(x+a+1)-ln(x+a)}{2\sqrt{x}}+ \frac{\sqrt{x}}{x+a+1} - \frac{\sqrt{x}}{x+a}$ $\int^{a+1}_{a} {\frac{ln(x+a+1)-ln(x+a)}{2\sqrt{x}} + \frac{\sqrt{x}}{x+a+1} - \frac{\sqrt{x}}{x+a}}\,dx =\sqrt{a+1}(ln(2a+2)-ln(2a+1))-\sqrt{a}(ln(2a+1)-ln(2a))$ Now $D$ is positive cause the double integral on the square $[a,a+1] [a,a+1]$ can be rewritten as ($T$ is the half triangle part of the square which lies above line $y=x$) $D= \int \int^{}_T {{\frac{y-x}{2\sqrt{x}(x+y)^2}}-{\frac{y-x}{2\sqrt{y}(x+y)^2}}}\,dy \,dx$ which is clearly positive.

Let's fix the idea in post #8. We must prove: $$6^{\sqrt{3}}4^{\sqrt{2}}>5^{\sqrt{3}+\sqrt{2}} $$Consider the function $f(x):=(5+x)^{\sqrt{3}}(5-x)^{\sqrt{2}}$. Put $a$ instead of $\sqrt{3}$ and $b$ instead of $\sqrt{2}$. We have: $$f(x)=(5+x)^{a-1}(5-x)^{b-1}(25-x^2)$$Let us denote $g(x):=(5+x)^{a-1}(5-x)^{b-1}$. Thus, we must prove $24g(1)>25g(0)$ or $24\left(g(1)-g(0)\right)>g(0)$. To estimate $g(1)-g(0)$ we use $g(1)-g(0)=\int_0^1 g'(t)\,dt$ and we need to estimate $g'(x)\,,\,x\in[0,1]$ (In fact, what was proved in #8 is that $g'(x)>0$ in that interval, that's g is increasing). We have: $$g'(x)=\left[(2-a-b)x+5(a-b)\right](5+x)^{a-2}(5-x)^{b-2}=\left[(2-a-b)x+5(a-b)\right]g_1(x) $$where $g_1(x):=(5+x)^{a-2}(5-x)^{b-2}$. It's easy to see $g_1(x)$ is increasing in $[0,1]$ but estimating it with $g_1(0)$ has not enough precision to prove what needed. So, we proceed further: $$g'_1(x)=\left[(4-a-b)x+5(a-b)\right] (5+x)^{a-3}(5-x)^{b-3} $$Since the RHS is increasing in $[0,1]$ (easy to see by noticing $4-a-b>0$ and taking the first derivative of $(5+x)^{a-3}(5-x)^{b-3}$ again) we have: $g'_1(x)>g'_1(0)=5(a-b)5^{a+b-6}$, hence: $$g_1(x)>g_1(0)+5(a-b)5^{a+b-6}x=5^{a+b-4}+5(a-b)5^{a+b-6}x $$Further, we get: $$g'(x)>\left[(2-a-b)x+5(a-b)\right]\left(5^{a+b-4}+5(a-b)5^{a+b-6}x \right)$$$$g(1)-g(0)=\int_0^1 g'(x)\,dx >\int_0^1 \left[(2-a-b)x+5(a-b)\right]\left(5^{a+b-4}+5(a-b)5^{a+b-6}x \right)\, dx $$Calculating the RHS, it follows: $$g(1)-g(0)>\left( 125(a-b)+\frac{25}{2}(a-b)^2-\frac{25}{2}(a+b-2)-\frac{5}{3}(a-b)(a+b-2) \right)5^{a+b-6} $$Thus, it's enough to prove $24\cdot \text{RHS }\geq g(0)=5^{a+b-2}$ or: $$125(a-b)+\frac{25}{2}(a-b)^2-\frac{25}{2}(a+b-2)-\frac{5}{3}(a-b)(a+b-2)\geq 5^4/24 $$which is easily checked (thou a hard bashing ).

I think post #14 is the best solution.

https://www.facebook.com/photo.php?fbid=1511949492260356&set=a.131305563658096.22284.100003359585981&type=3&theater

math90 wrote: I think post #14 is the best solution. Yeah, I completely agree. I call that type of proofs: "the moments of truth". It not only proves the claim, but shows WHY it's true, the big picture. @estoyanovvd: Your link is broken. The fact you see something in FB doesn't mean somebody else also does! Please, provide another one, or just copy it here. Btw, the problem was proposed by Nikolai Nikolov.

Lamp909 wrote: Prove that $$(\frac{6}{5})^{\sqrt{3}}>(\frac{5}{4})^{\sqrt{2}}$$. Proof of lys :

Nice, but how do you prove $2x-(x+2)\ln(1+x)<0$ ?. It's true, but one should use the expansion of $\ln(1+x)$ up to $x^4$, that's $\ln(1+x)>x-x^2/2+x^3/3-x^4/4$. But anyway it saves some calculations as in #6. The official solution also uses many calculations. It estimates $(1+x)^{\alpha}$ using Taylor expansion of it up to $x^3$.

Because $$\left(\ln(1+x)-\frac{2x}{x+2}\right)'=\frac{1}{1+x}-\frac{4}{(x+2)^2}=\frac{x^2}{(x+1)(x+2)^2}\geq0.$$

My generalization and solution by Kunihiko Chikaya.

Prove that $$\left(\frac{6}{5}\right)^{\sqrt{3}}>\left(\frac{5}{4}\right)^{\sqrt{2}}.$$ A bit off topic, but the formatting should be like this

Taking log on both sides we get to show that the result is true. That is √3(log6–log5)>√2(log5–log4) Or √3(log2+log3–log5)>√2(log5–2log2) Or since log2=.3010,log3=.4771 log5=log10–log2=.6989(rounded off to 4 significant figures). Then lhs comes out to be. .1371.... And lhs comes our to be. .1370...... Which is obviously true.

estoyanovvd wrote: My generalization and solution by Kunihiko Chikaya. @estoyanov: That solution, you provided, unfortunately is wrong! Otherwise, a nice idea indeed. So, let me explain. We have the function $f(n):= (n-1)\ln\frac{n}{n-1}$ and want to prove $f(n+1)>f(n)$. Naturally, by trying to prove $f(x)$ is increasing, that's $f'(x)>0$. So far, so good. It's equivalent to: $$(*)\,\,\,\,\,\,\,\,\,\, \ln\frac{x}{x-1}>\frac{2(x-3)}{x(x-1)}\,,\, x\geq 5 $$which is OK. But, I can't understand the next two lines, they are very wrong, besides why should we use AM-GM, etc., when all the terms are positive?! Anyway, it even cannot be fixed, because $(*)$ is not true by itself. Use a calculator and check it for $x=7$ for example. @Wolowizard: I don't agree at all. That hard calculations are indeed too much, but there are also good and instructive solutions. @paragdey01: You use a calculator to calculate that log's which is prohibited (on the competition).

paragdey01 wrote: Taking log on both sides we get to show that the result is true. That is √3(log6–log5)>√2(log5–log4) Or √3(log2+log3–log5)>√2(log5–2log2) Or since log2=.3010,log3=.4771 log5=log10–log2=.6989(rounded off to 4 significant figures). Then lhs comes out to be. .1371.... And lhs comes our to be. .1370...... Which is obviously true. Your approximations are too close to call.

Lamp909 wrote: Prove that \[ \left(\frac{6}{5}\right)^{\sqrt{3}}>\left(\frac{5}{4}\right)^{\sqrt{2}}. \]

A solution carried out using an abacus, no paper, and no calculus. Brought to you by Twitch Solves ISL: Because $2.45^2 = 6.0025$, it follows $\sqrt{6} < 2.45$. So, it is enough to show that \[ \left( \frac 65 \right)^3 > \left( \frac 54 \right)^{2.45} \]which is true if and only if \[ 3^{60} \cdot 2^{267} > 10^{109}. \]Using abacus magic, we compute \begin{align*} 2^{267} &= \frac 18 \cdot 1024^{27} = \frac 18 \cdot \left( 1048576 \cdot 1024 \right)^9 \\ &= \frac 18 \cdot \left( 1073741824 \right)^9 \\ &> \frac 18 \cdot \left( 10737418 \right)^9 \cdot 10^{18} \\ &> \frac 18 \cdot \left( 115292145300000 \cdot 10737418 \right)^3 \cdot 10^{18} \\ &> \frac 18 \cdot \left( 1237939956 \cdot 10^7 \right)^3 \cdot 10^{33} \\ &= 618969978^3 \cdot 10^{54} \end{align*}The main calculations are shown here. 10737418 1152921453 75161926 8070450171 32212254 3458764359 75161926 8070450171 42949672 4611685812 10737418 1152921453 85899344 9223371624 --------------- ----------------- 115292145306724 12379399562028354 On the other hand, we have \[ 3^{60} = 59049^6 = \left( 3486784401 \right)^3 \]This means that we need only show \[ \left( 3486784401 \cdot 618969978 \right)^3 > 10^{49} \iff 3486784401 \cdot 618969978 \ge \sqrt[3]{10} \cdot 10^{16} \]In fact, we will show that \[ 34867844 \cdot 618969978 \ge \sqrt[3]{10} \cdot 10^{14}. \]Computing the first bits of the partial products, the left hand side is more than $2.1582 \cdot 10^{15}$. On the other hand \begin{align*} 2.158 \cdot 2.158 &= 4.656964 \\ 4.656964 \cdot 2.15 &= 10.0124726 > 10. \end{align*}This completes the proof.

It can be done by looking of a close approximate to √6 and using binomial expansion to estimate exponent values. I've posted my solution in Youtube. https://youtu.be/UnUTv1TS-ls

Equivalant to showing that $1.2^{3}>1.25^{\sqrt{6}}$ Since $\sqrt{6}<2.45$, it is sufficient to show that $(\frac{6}{5})^3>(\frac{5}{4})^{2.45}$, which is true if and only if $3^{60} \cdot 2^{267} >10^{109}$. Now it is not too hard hopefully?? Unless it is which is likely because this is a P3. But also pls no Hopefully, the LHS has $\geq$the number of digits as the RHS. Otherwise cri Claim. The LHS has $\geq$110 digits. Proof. $2^{10}=1024$, and it follows that $2^{260}$ has $79$ digits. It also follows that $2^{267}$ has $81$ digits. Now, it is a matter of deducing the amount of digits that $3^{60}$ has. Note that $3^{60}=9^{30}=(10-1)^{30}$ For binomial theorem and hockey stick identity reasons this results in $3^{60}$ having $29$ digits. This means that the claim is true because $29+81=110$ and $3^{60}\cdot2^{267}>10^{109}$ Since this implies the problem, the proof is complete.

$2^{267}$ having 81 digits means $2^{267}>10^{80}$. Similarly, $3^{60}$ having 29 digits means $3^{60}>10^{28}$. Therefore, it could only prove that: $$3^{60}\cdot 2^{267}>10^{80+28}=10^{108}$$not $ 10^{109}$.

I think this problem is good Since it can be proved without calculus technique or binomial expansion. or many computations. $(\frac{6}{5})^{\sqrt{3}} > (\frac{5}{4})^{\sqrt{2}}$ $\;\;\Leftrightarrow (\frac{6}{5})^{\sqrt{3/2}} > \frac{5}{4}$ $\;\;\Leftarrow (\frac{2k+2}{2k+1})^{\sqrt{1+1/k}} > \frac{2k+1}{2k}$ for $k>0$ Substitute $\sqrt{1+1/k}$ with $s$, then $k = \frac{1}{s^2-1}$ and $\;\;\Leftrightarrow (\frac{2+2(s^2-1)}{2+(s^2-1)})^s > \frac{2+(s^2-1)}{2}$ for $s>1$ $\;\;\Leftrightarrow (\frac{2s^2}{s^2+1})^s > \frac{s^2+1}{2}$ $\;\;\Leftrightarrow s^{2s} > (\frac{s^2+1}{2})^{s+1} \;\;\Leftrightarrow s^{\frac{2s}{s+1}-1} > {\frac{s^2+1}{2s}}$ $\;\;\Leftrightarrow s^{\frac{s-1}{s+1}} > {\frac{s^2+1}{2s}} = \frac{(s+1)^2+(s-1)^2}{(s+1)^2-(s-1)^2} = \frac{1+(\frac{s-1}{s+1})^2}{1-(\frac{s-1}{s+1})^2} $ Now we partly substitute $\frac{s-1}{s+1}$ with $x$ and $t$ and introduce a function $f$ $\;\;\Leftrightarrow s^x > \frac{1+tx}{1-tx} = f(x)$ Our plan is to show the inequality $s^x > f(x)$ holds for $0<x<1$. (treat $s>1$ and $0<t<1$ as constants, so the inequality is just about $x$.) where $f$ is a function defined by $f(x) = \frac{1+tx}{1-tx}$ for $0 \leq x \leq 1$. Note that $s^0 = f(0), s^1 = f(1).$ We will show that for any $0 \leq a \neq b \leq 1$, the following is true : $s^a \geq f(a)$ and $s^b \geq f(b) \Rightarrow s^\frac{a+b}{2} > f(\frac{a+b}{2})$ ...(*) It suffices to prove that $ f(a)f(b) > f(\frac{a+b}{2})^2 $ that is $\frac{1+ta}{1-ta} \frac{1+tb}{1-tb}$ > $(\frac{1+t\frac{a+b}{2}}{1-t\frac{a+b}{2}})^2$ Substitute $ta, tb$ with $A,B$ (note that $A \neq B$) then we get $\;\;\Leftrightarrow$ $ \frac{1+A}{1-A} \frac{1+B}{1-B} = \frac{1+AB+(A+B)}{1+AB-(A+B)} = \frac{1+(A+B)/(1+AB)}{1-(A+B)/(1+AB)}$ $> (\frac{1+\frac{A+B}{2}}{1-\frac{A+B}{2}})^2 = \frac{1+(\frac{A+B}{2})^2+(A+B)}{1+(\frac{A+B}{2})^2-(A+B)} = \frac{1+(A+B)/(1+(\frac{A+B}{2})^2)}{1-(A+B)/(1+(\frac{A+B}{2})^2)}$ $\;\;\Leftrightarrow$ $(A+B)/(1+AB)$ > $(A+B)/({1+(\frac{A+B}{2})^2})$ $\;\;\Leftrightarrow$ $ 1+AB < 1+(\frac{A+B}{2})^2 $ $\;\;\Leftrightarrow$ $ A \neq B$ We showed (*). It means $s^x > f(x)$ is true for rational numbers with a form $\frac{m}{2^n} (0<m<2^n)$. Since $s^x$ and $f(x)$ are continuous function, we get $s^x \geq f(x)$ for all real $0<x<1$. (In fact this is the most non-elementary step.) To remove $=$ from $\geq$, we use (*) again : Any $0<x<1$ is a form of $\frac{a+b}{2}$ for some $0 \leq a \neq b \leq 1$. So by (*) we get $s^x > f(x)$ for all real $0<x<1$. The answer follows from $s^t > f(t)$.

Taking logarithms yields the equivalent $\ln(1 + \frac{1}{4})/\ln(1+\frac{1}{5}) > \frac{\sqrt{6}}{2}$, i.e. $$ \frac{\ln\left(1+\frac{1}{4}\right)\sqrt{\left(1+\frac{1}{4}\right)}}{\frac{1}{4}} < \frac{\ln\left(1+\frac{1}{5}\right)\sqrt{\left(1+\frac{1}{5}\right)}}{\frac{1}{5}}.$$Hence it suffices to show that $f(x) = \frac{\ln(x+1)\sqrt{x+1}}{x}$ is a decreasing function for $x>0$. We compute $$ f'(x) = \frac{2x-(x+2)\ln(x+1)}{2x^2\sqrt{x+1}} \mbox{ and } \left(\frac{2x}{x+2} - \ln(x+1)\right)'= -\frac{x^2}{(x+1)(x+2)^2} < 0 $$so the numerator of $f'$ is strictly decreasing. This numerator equals $0$ for $x=0$, and the denominator of $f'$ is positive for $x>0$ -- therefore $f'(x) < 0$ for $x>0$ and we are done.