Well it's nice to have an inequality problem anyways...

First impression-do something on $d+c+b+a$, then $c+b+a$, then $b+a$, then $a$ Althoguh probably homogenize first with $(a+b+c+d)^2$?? I probably will try this one more later because it looks interesting

By Weighted AM-GM $a^2 + b^2 + c^2 + d^2 \geq a^ab^bc^cd^d$. It remains to show that\[(a + 2b + 3c + 4d)(a^2 + b^2 + c^2 + d^2) \leq (a + b + c + d)^3.\]I don't know the details but this can probably be done with expansion. Edit: Yeah @below seems to confirm that expansions works.

Click to reveal hidden text

Also you can just log everything and then weighted jensen should work

Small rant, but honestly I was pretty disappointed when I saw this ineq :/ Like, why :/

jj_ca888 wrote: By Weighted AM-GM $a^2 + b^2 + c^2 + d^2 \geq a^ab^bc^cd^d$. It remains to show that\[(a + 2b + 3c + 4d)(a^2 + b^2 + c^2 + d^2) \leq (a + b + c + d)^3.\]I don't know the details but this can probably be done with expansion. Bufallo method finishes this. $a=d+x+y+z, b=d+x+y, c=d+x, d=d$. Then expand. All coeffs on RHS are more than corresponding on LHS.

The expansion in post #4 does seem to work (which means the problem is just two unrelated problems put together ). By weighted AM-GM we have \[ a^a b^b c^c d^d \le \sum_{\text{cyc}} \frac{a}{a+b+c+d} \cdot a = a^2+b^2+c^2+d^2. \]Consequently, it is enough to prove that \[ (a^2+b^2+c^2+d^2)(a+2b+3c+4d) \le 1 = (a+b+c+d)^3. \]Expand both sides to get \[ \begin{array}{cccc} +a^3 &+ b^2a &+ c^2a & +d^2a \\ +2a^2b &+ 2b^3 &+ 2bc^2 & +2d^2b \\ +3a^2c & + 3b^2c & + 3c^3 & + 3d^2c \\ +4a^2d &+ 4b^2d & + 4c^2d & + 4d^3 \end{array} < \begin{array}{cccc} +a^3 &+ 3b^2a &+ 3c^2a & +3d^2a \\ +3a^2b &+ b^3 &+ 3bc^2 & +3d^2b \\ +3a^2c &+ 3b^2c &+ c^3 &+ 3d^2c \\ +3a^2d &+ 3b^2d &+ 3c^2d &+ d^3 \\ +6abc &+ 6bcd &+ 6cda &+ 6dab \end{array} \]In other words, we need to prove that \[ \begin{array}{cccc} & && \\ &+ b^3 & & \\ & & +2c^3 & \\ +a^2d &+ b^2d & + c^2d & + 3d^3 \\ \end{array} < \begin{array}{cccc} &+ 2b^2a &+ 2c^2a & +2d^2a \\ +a^2b & &+ bc^2 & +d^2b \\ &&& \\ &&& \\ +6abc &+ 6bcd &+ 6cda &+ 6dab \end{array} \]This follows since \begin{align*} 2b^2a &\ge b^3 + b^2d \\ 2c^2a &\ge 2c^3 \\ 2d^2a &\ge 2d^3 \\ a^2b &\ge a^2d \\ bc^2 &\ge c^2d \\ d^2b &\ge d^3 \end{align*}and $6(abc+bcd+cda+dab) > 0$.

naman12 wrote: The real numbers $a, b, c, d$ are such that $a\geq b\geq c\geq d>0$ and $a+b+c+d=1$. Prove that \[(a+2b+3c+4d)a^ab^bc^cd^d<1\] By weightet AM-GM $LHS <= (a+2b+3c+4d)(a^2+b^2+c^2+d^2)$. Hence, it suffices $(a+2b+3c+4d)(a^2+b^2+c^2+d^2)<(a+b+c+d)(a^2+b^2+c^2+d^2+2\sum{ab})$ or $(b+2c+3d)(a^2+b^2+c^2+d^2) <(a+b+c+d)(2\sum{ab})$. $LHS \le (b+2c+3d)a(a+b+c+d)$. Hence, it suffices to prove that $ab+2ac+3ad \le 2\sum{ab}$ or that $ad \le ab+ something$. That's obvious.

The weighted AM-GM part working feels really artificial.

naman12 wrote: The real numbers $a, b, c, d$ are such that $a\geq b\geq c\geq d>0$ and $a+b+c+d=1$. Prove that \[(a+2b+3c+4d)a^ab^bc^cd^d<1\] When the PSC tries not to make it CCCCCC, this happens.

Taking log: $log(a+2b+3c+4d)+a\cdot log(a)+b \cdot log(b) + c \cdot log(c)+d \cdot log(d) \leq(1+a+b+c+d)log \left( \frac{a+2b+3c+4d+a^2+b^2+c^2+d^2}{1+a+b+c+d} \right) <1$ by weighted jensen since $log$ is concave.

Redacted

I really wonder what the algebra shortlist for IMO 2020 looks like. First by weighted AM-GM, we find $a^2+b^2+c^2+d^2 \geq a^ab^bc^cd^d,$ so we rearrange the problem to $$(a^2+b^2+c^2+d^2)(a+2b+3c+4d) < 1 = (a+b+c+d)^3.$$Now proceed similarly to v_Enhance.

Is there any nicer solution than bashing it out at the end? Doesn’t seem like it so far...

See also here:

circlethm wrote: Is there any nicer solution than bashing it out at the end? Doesn’t seem like it so far... see post #10

Here is a new clean/direct solution. How is everyone so good at expansion Let $t = a+2b+3c+4d$. By weighted AM-GM, \[ \left(a^{a+2b+2c+2d} b^{c+2d}\right)^{1/t} \leq \frac{a(a+2b+2c+2d) + b(c+2d)}{t} < \frac{(a+b+c+d)^2}{t} = \frac 1t, \]so to finish it suffices to show that \[ a^{a+2b+2c+2d} b^{c+2d}\geq a^{at} b^{bt} c^{ct} d^{dt}. \]Since $a\geq b \geq c \geq d$, we just need to show that $(a+2b+2c+2d,c+2d,0,0)$ majorizes $(at,bt,ct,dt)$. But this is equivalent to \[ a+2b+2c+2d = 2-a \geq at. \]But $t \leq a+3(b+c+d) = a+3(1-a) = 3-2a$, so we just need \[ 2-a \geq 3a-2a^2 \iff 2(a-1)^2 \geq 0, \]which is clearly true!

A bit of motivation: Basically the best way to deal with expressions of the form $x^x$ is to use $x^x = e^{xlogx}$. Now using Jensen, and the condition $a+b+c+d=1$ we obtain $a \log{a}+b \log{b}+c \log{c}+d \log{d} \leq \log{(a^2+b^2+c^2+d^2)}$. Now note that $$a^a \cdot b^b \cdot c^c \cdot d^d = e^{a \log{a}+b \log{b}+c \log{c}+d \log{d}} \leq e^{log(a^2+b^2+c^2+d^2)} = (a^2+b^2+c^2+d^2)$$ Now just homogenize to see that it is enough to prove $$(a+2b+3c+4d)(a^2+b^2+c^2+d^2) < (a+b+c+d)^3 $$ But this is just direct expansion, and a bit of experience is enough to finish it. There is basically only one way to solve the problem, and in my opinion it isn't something creative or new.

Generalization 1 Let $a,b,c,d$ be reals ($a\geq b\geq c\geq d>0$) such that $\alpha\geq \theta\geq \beta\geq \lambda>0$ and $$6\lambda \geq \alpha +\beta \geq 2\theta$$inequalities hold. Then prove that $$\left(\lambda a+\beta b+\theta c+\alpha d\right)a^ab^bc^cd^d<\lambda$$

sobad . By Weighted AM-GM we have $a^ab^bc^cd^d \le a^2+b^2+c^2+d^2.$ Then we want to show that $(a+b+c+d)^3-(a+2b+3c+4d)(a^2+b^2+c^2+d^2)>0,$ but upon expanding this becomes $a^2(b-d)+b^2(2a-b-d)+c^2(2a+b-2c-d)+d^2(2a+b-d)+6(abc+abd+acd+bcd)>0$ which is clear.

Steve12345 wrote: Taking log: $log(a+2b+3c+4d)+a\cdot log(a)+b \cdot log(b) + c \cdot log(c)+d \cdot log(d) \leq(1+a+b+c+d)log \left( \frac{a+2b+3c+4d+a^2+b^2+c^2+d^2}{1+a+b+c+d} \right) <1$ by weighted jensen since $log$ is concave. Doesn't the $R.H.S$ become $0$ after taking $log$? Then the problems becomes proving that our $log$ is negative, or in other words that the expression inside is less than $1$ which I'm not sure if it is true or not.

By am-gm inequality's weighted form It is very easy to get $a^2+b^2+c^2+d^2 \geq a^ab^bc^cd^d$ So it remains to prove $(a+2b+3c+4d)(a^2+b^2+c^2+d^2)<1$ Actually $1=(a+b+c+d)^x$ And we can easiyl deduce that $x$ can be 3. So,rest is boring calculation. Which solves problem.

mathmax12 wrote: Claim 1: $a^2+b^2+c^2+d^2 \ge a^a \cdot b^b\cdot c^c\cdot d^d.$ Proof: Follows, by Weighted, AM-GM. Now, all we need to show, is that $(a^2+b^2+c^2+d^2)(a+2b+3c+4d)$, this is maximized, when $a=b=c=d$, so, $(a+2b+3c+4d)(a^2+b^2+c^2+d^2)=40a^3=40\cdot\frac{1}{64}=\frac{5}{8}<1$, so, we are done. YaoAOPS wrote: Note that \[ a^ab^bc^cd^d \le a^2 + b^2 + c^2 + d^2 \]follows by weighted AM-GM. It thus remains to show that \[ (a + 2b + 3c + 4d)(a^2 + b^2 + c^2 + d^2) \le 1 \] Claim: The expression is maximized when all variables are equal. Proof. Suppose not, and that the variables can be divided into equal groups. Then consider perturbing one group by $\varepsilon$ and the other by $c\varepsilon$ such that $a + b + c + d = 1$ still holds. By taking both a positive $\varepsilon$ and $-\varepsilon$, and letting $\varepsilon$ go to $0$, this gives two possible new values for the expression, one of which must be larger. $\blacksquare$ Finally, when $a = b = c = d$ this gives the value of $\frac{5}{8} < 1$. I believe both solutions are incorrect. I’m not sure what you’re doing with the “perturbing”, but the maximum value is not when $a=b=c=d$, but it’s approached when $a \to 1$ and $b, c, d \to 0$.

AngeloChu wrote: Like literally everyone else said, we get $a^ab^bc^cd^d \le a^2+b^2+c^2+d^2$ from weighted AM-GM We need $(a+2b+3c+4d)(a^2+b^2+c^2+d^2) \le 1$ From Cauchy's, we get $a^2+b^2+c^2+d^2 \le \frac{(a+b+c+d)^2}{4}=\frac{1}{4}$ Now we have $\frac{(a+2b+3c+4d)}{4} \le 1$ since that is what we get by plugging in $a^2+b^2+c^2+d^2 \le \frac{1}{4}$

We can simplify to $a+2b+3c+4d \le 4$ with equality at $(a,b,c,d)=(0,0,0,1)$ Click to reveal hidden text

However, it is impossible since $a\ge b\ge c\ge d>0$ Therefore, $a+2b+3c+4d < 4$ Reversing the steps, we get: $$\frac{(a+2b+3c+4d)}{4} < 1$$$$(a+2b+3c+4d)(a^2+b^2+c^2+d^2) < 1$$$$(a+2b+3c+4d)a^ab^bc^cd^d<1$$And our inequality holds. This solution is also wrong since Titu's lemma actually says that $$a^2+b^2+c^2+d^2 \ge \frac{(a+b+c+d)^2}{4}=\frac{1}{4}$$

Weighted AM-GM + critical values. Use weighted AM-GM, we get $a^ab^bc^cd^d \le a^2 + b^2 + c^2 + d^2$ and we have \[(a+2b+3c+4d)(a^2+b^2+c^2+d^2) \le 1\]Know that $b \le d$, then $(a + 3b + 3c + 3d)(a^2 + b^2 + c^2 + d^2) < 1$ $a + 3b + 3c + 3d$ can be written into $3 - 2a$ since $a + b + c + d = 1$ and we have \[(3-2a)(a^2+b^2+c^2+d^2) \le 1\] Assume $a = x$, $b=c=d=y$, then $x+3y = 1$, $y = (1-x)/3$, notice that $x > y$. Substitute $y$ back to $a^2 + b^2 + c^2 + d^2$, we get $a^2 + b^2 + c^2 + d^2 = (4x^2 - 2x + 1)/3$ Then, we get \[(3 - 2x)(4x^2 - 2x + 1)/3 < 1\] We can use critical values to find the maximum value of $(3 - 2x)(4x^2 - 2x + 1)/3$. In this cases we have maximum value $49/81$ with $x = 1/3$ and $y = 2/9$ Since $49/81 < 1$, $(a + 3b + 3c + 3d)(a^2 + b^2 + c^2 + d^2) < 1$, and $(a + 2b + 3c + 4d)(a^2 + b^2 + c^2 + d^2) < 1$, and we have \[(a+2b+3c+4d)a^ab^bc^cd^d<1\]

Opening remark: Why does this even work By Weighted AM-GM, we have $$a^a b^b c^c d^d \le a^2 + b^2 + c^2 + d^2.$$So it suffices to prove that $(a+2b+3c+4d)(a^2 + b^2 + c^2 + d^2) < (a+b+c+d)^3$, which is true by \textit{coughs} expanding. Closing remark: In particular, the inequality should be equivalent to $$a^2b + 2ab^2 + 2ac^2 + 2ad^2 + bc^2 + bd^2+6abc +6bcd+6cda+6dab$$$$> b^3 + 2c^3 + 3d^3 + a^2d + b^2d + c^2d,$$assuming my expansion is correct. This can be solved by "cancelling" terms from the right hand side with larger ones from the left hand side.

bashee_wang wrote: plagueis wrote: This is a very sad problem Why this is a "sad" problem? For something which is P2 IN THE IMO you would expect it to be challenging.But no,even most 9th grader can solve this with just a little weighted AM-GM and direct expansion.

IceyCold wrote: bashee_wang wrote: plagueis wrote: This is a very sad problem Why this is a "sad" problem? For something which is P2 IN THE IMO you would expect it to be challenging.But no,even most 9th grader can solve this with just a little weighted AM-GM and direct expansion. I mean the fact that expansion works is really hard to see, especially with the thoughts of "this is the IMO" in the back of your head.

I mean it may be hard to see due to the fact that this is the IMO.But if 9th graders can solve this with no issues then why throw it in the IMO in the first place?I might be wrong here though.

How is this problem real. WAMGM obviously reduces this to $(a+2b+3c+4d)(a^2+b^2+c^2+d^2) < 1$ or homogenizing it $$(a+2b+3c+4d)(a^2+b^2+c^2+d^2) <(a+b+c+d)^3$$Which after expansion and mass canceling reduces to the sum of $$a^2d \leq a^2b$$$$c^2d \leq bc^2$$$$b^3 \leq ab^2$$$$d^3\leq bd^2$$$$d^3 \leq bd^2$$$$2c^3 \leq 2ac^2$$$$2d^3 \leq 2ad^2$$$$b^2d \leq ab^2$$$$6(abc+abd+acd+bcd) > 0$$done.

Steve12345 wrote: Taking log: $log(a+2b+3c+4d)+a\cdot log(a)+b \cdot log(b) + c \cdot log(c)+d \cdot log(d) \leq(1+a+b+c+d)log \left( \frac{a+2b+3c+4d+a^2+b^2+c^2+d^2}{1+a+b+c+d} \right) <1$ by weighted jensen since $log$ is concave. By taking log on both sides, should the RHS be 0 not 1?

The key is to use AM-GM in the following way: $$a^{a}b^{b}c^{c}d^{d} \le \left(\frac{a^{2}+b^{2}+c^{2}+d^{2}}{a+b+c+d}\right)^{a+b+c+d} = a^{2}+b^{2}+c^{2}+d^{2}.$$So, we need to show that $$(a+2b+3c+4d)(a^{2}+b^{2}+c^{2}+d^{2}) \le 1.$$Now, we substitute $1 = (a+b+c+d)^{3},$ we get $$(a+2b+3c+4d)(a^{2}+b^{2}+c^{2}+d^{2}) \le (a+b+c+d)^{3}.$$Now, we expand, and then compare terms, using $a \le b \le c \le d.$

Funny story: I solved this problem during a Socratic seminar in English. i started the expansion as this guy began yapping. By the time I solved the problem, he still wasn't done.

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