Ahiles wrote: Find all real numbers $ a,b,c,d$ such that $ \{\begin{array}{cc}a + b + c + d = 20 \\ ab + ac + ad + bc + bd + cd = 150 \end{array}$ $ ab + ac + ad + bc + bd + cd = 150$ > $ 400=150*\frac{8}{3}=\frac{8}{3}(ab + ac + ad + bc + bd + cd)=2(ab + ac + ad + bc + bd + cd)+\frac{1}{3}(2ab + 2ac + 2ad + 2bc + 2bd + 2cd)\leq2(ab + ac + ad + bc + bd + cd)+\frac{1}{3}(a^{2}+b^{2}+a^{2}+c^{2}+a^{2}+d^{2}+b^{2}+c^{2}+b^{2}+d^{2}+c^{2}+d^{2})=2(ab + ac + ad + bc + bd + cd)+a^{2}+b^{2}+c^{2}+d^{2}=(a+b+c+d)^{2}=20^{2}=400$ > $ a=b=c=d=5$

Let $ S = ab + ac + ad + bc + bd + cd = 150$. We know that $ (a+b+c+d)^2 = a^2 + b^2 + c^2 + d^2 + 2S$, so $ a^2 + b^2 + c^2 + d^2 = 400 - 300 = 100$. So we have to find all real numbers $ a, b, c, d$ such that (1) $ a + b + c + d = 20$, and (2) $ a^2 + b^2 + c^2 + d^2 = 100$. By AM-QM inequality we have that $ \frac {a+b+c+d}{4} \le \sqrt {\frac {a^2 + b^2 + c^2 + d^2}{4}}$, with equality if and only if $ a=b=c=d$. But these expressions are in fact equal, as they both equal 5. So the only solution is $ a=b=c=d=5$

uglysolutions wrote: Let $ S = ab + ac + ad + bc + bd + cd = 150$. We know that $ (a + b + c + d)^2 = a^2 + b^2 + c^2 + d^2 + 2S$, so $ a^2 + b^2 + c^2 + d^2 = 400 - 300 = 100$. So we have to find all real numbers $ a, b, c, d$ such that (1) $ a + b + c + d = 20$, and (2) $ a^2 + b^2 + c^2 + d^2 = 100$. By AM-QM inequality we have that $ \frac {a + b + c + d}{4} \le \sqrt {\frac {a^2 + b^2 + c^2 + d^2}{4}}$, with equality if and only if $ a = b = c = d$. But these expressions are in fact equal, as they both equal 5. So the only solution is $ a = b = c = d = 5$ Can we use AM-QM for negative numbers?

Johan Gunardi wrote: uglysolutions wrote: By AM-QM inequality we have that $ \frac {a + b + c + d}{4} \le \sqrt {\frac {a^2 + b^2 + c^2 + d^2}{4}}$, with equality if and only if $ a = b = c = d$. But these expressions are in fact equal, as they both equal 5. So the only solution is $ a = b = c = d = 5$ Can we use AM-QM for negative numbers? No, you can't use AM-QM for negative numbers. But the gap in the solution is easily repaired: If all four numbers are non-negative, then the above argument works. If one of the four numbers is negative, then $ \frac {a + b + c + d}{4} < \frac {|a| + |b| + |c| + |d|}{4} \le \sqrt {\frac {a^2 + b^2 + c^2 + d^2}{4}}$ and you get a contradiction.

That is really triviality

With the given equations, we have $ \sum (a - 5)^2 = \sum a^2 - 10\sum ab + 100 = 0$ (the summations are cyclic), that is, $ a = b = c = d = 5$.

I suggest the following generalized problem: Find all real numbers $ a_{1},a_{2},...,a_{n}$ which satisfy the conditions $ \sum_{i=1}^{n}a_{i}=k>0$ and $ \sum_{i,j=1, i\ne j}^{n}a_{i}a_{j}=\frac{k^{2}(n-1)}{n}$.

Is its solution similar to ?

Yes! The main idea is that $ (\sum_{i=1}^{n}a_{i})^2=\sum_{i,j=1,i\ne j}^{n}a_{i}a_{j}+\sum_{i=1}^{n}a_{i}^{2}$. The final answer is $ a_{1},a_{2},\cdots,a_{n}=\frac{k}{n}$

with this generalization we can use a genaralization of the inequality FOURRIER proposed

No, I think AM-QM can works for negative numbers in the case $ p=2k$.. Because it's similar with Cauchy Schwarz Ineqauality : $ 4(a^2+b^2+c^2+d^2) \ge (a+b+c+d)^2$ When (at least) one of the numbers is negative, the value will be smaller, so the ineq is still true..

yes, Ronald, you are right $ \frac{a+b+c+d}{4} \le \sqrt{\frac{a^2+b^2+c^2+d^2}{4}}$ if one (or more) of $ a,b,c,d$ is negative, then the value of $ LHS$ will decrease and won't ruin the system... boxedexe's solution is good.. $ (a-5)^2+(b-5)^2+(c-5)^2+(d-5)^2=0$

Ahiles wrote: Find all real numbers $ a,b,c,d$ such that \[ \left\{\begin{array}{cc}a + b + c + d = 20, \\ ab + ac + ad + bc + bd + cd = 150. \end{array} \right. \] $ 3(a^2+b^2+c^2+d^2)=300$ $ 2(ab + ac + ad + bc + bd + cd)=300$ so $ 3(a^2+b^2+c^2+d^2)-2(ab + ac + ad + bc + bd + cd)=0$ $ (a-b)^2+(a-c)^2+(a-d)^2+(b-c)^2+(b-d)^2+(c-d)^2=0$ so $ a=b=c=d=5$

Suppose the problem was to solve for positive reals, it can be solved without any equation systems but by trivial AM- GM inequality. a+b+c+d>= 4(abcd)^1/4 >* and ab+ac+ad+bc+bd+cd>=6 (abcd)^1/2 > ** which yields abcd=5 ^ 4 where AM=GM implying the fact that a=b=c=d=5.

You can also use the idea from the MMO '11, and look at it as a polynomial. After we get $a^2+b^2+c^2=100-d^2$ and $a+b+c=20-d$, using Cauchy we get: $(a^2+b^2+c^2)\cdot (1+1+1)\ge (a+b+c)^2$ $(100-d^2)\cdot 3\ge (20-d)^2$ $(d-5)^2\le 0$ for an arbitrary d, which implies $a=b=c=d=5$.

Similar Problem: See here 1978 USAMO Problem 1

The numbers $a,b,c,d$ are the real roots of the polynomial $P(x)=x^4-20x^3+150x^2+mx+n$, for some reals $m,n$. If they are not equal, then $P'(x)$ has three non-equal roots and hence $P''(x)$ has two non-equal roots. But $P''(x)=12(x^2-10x+25)$, whose roots are equal to $5$. We deduce that $a=b=c=d=5$. I used a consequence of Rolle's Theorem

enescu wrote: The numbers $a,b,c,d$ are the real roots of the polynomial $P(x)=x^4-20x^3+150x^2+mx+n$, for some reals $m,n$. Let the quartic function $y=P(x)$ touch the line $y=l(x)$, Rewrite $P(x)=(x^2-10x+p)^2+\cdots$, yielding $p=25$ by observing the coefficient of $x^2$, thus We can write $P(x)=(x-5)^4+(m-500)x+n$, since $P(x)$ is a polynomial the fourth degree, therefore $P(x)=0\Longleftrightarrow x=5$, that is to say, $(m-500)x+n\equiv0\Longleftrightarrow m=500,\ n=0$.

enescu wrote: The numbers $a,b,c,d$ are the real roots of the polynomial $P(x)=x^4-20x^3+150x^2+mx+n$, for some reals $m,n$. If they are not equal, then $P'(x)$ has three non-equal roots and hence $P''(x)$ has two non-equal roots. But $P''(x)=12(x^2-10x+25)$, whose roots are equal to $5$. We deduce that $a=b=c=d=5$. I used a consequence of Rolle's Theorem A slight imprecision. If $a,b,c,d$ are pairwise distinct, or if two of them are of equal value $v$ and the other two are either of distinct values different from $v$, or of equal value different from $v$, then indeed $P'(x)$ has three non-equal real roots. But if three of them are equal, and the fourth is of a different value, then $P'(x)$ has two equal roots, and a third of a different value, still however enough for $P''(x)$ to have two non-equal real roots - leading to the above contradiction.

kunny wrote: enescu wrote: The numbers $a,b,c,d$ are the real roots of the polynomial $P(x)=x^4-20x^3+150x^2+mx+n$, for some reals $m,n$. Let the quartic function $y=P(x)$ touch the line $y=l(x)$, Rewrite $P(x)=(x^2-10x+p)^2+\cdots$, yielding $p=25$ by observing the coefficient of $x^2$, thus We can write $P(x)=(x-5)^4+(m-500)x+n$, since $P(x)$ is a polynomial the fourth degree, therefore $P(x)=0\Longleftrightarrow x=5$, that is to say, $(m-500)x+n\equiv0\Longleftrightarrow m=500,\ n=0$. Slightly inintelligible (what is that "line" $y=l(x)$, and what role does it play?). Also slightly wrong; in fact $P(x) = (x-5)^4 + (m-500)x + (n-625)$. Now the follow-up (which I still fail to fully understand ...) at least reaches the correct values $m=500$, $n= 625$.

Thank you for the correction in replace of me, mavropnevma.

Solution : $(a+b+c+d)^2-2*(ab+ac+ad+bc+bd+cd)=a^2+b^2+c^2+d^2=100$ , $a^2+b^2+c^2+d^2\ge ab+bc+ca+ab$ , like the same $a^2+c^2+b^2+d^2\ge ac+cb+bd+da$ , $a^2+b^2+d^2+c^2\ge ab+bd+dc+ca$ , so $300=3*(a^2+b^2+c^2+d^2)\ge 2*(ab+ac+ad+bc+bd+cd)=300$ , so $a=b=c=d=5$ . done

mavropnevma wrote: enescu wrote: The numbers $a,b,c,d$ are the real roots of the polynomial $P(x)=x^4-20x^3+150x^2+mx+n$, for some reals $m,n$. If they are not equal, then $P'(x)$ has three non-equal roots and hence $P''(x)$ has two non-equal roots. But $P''(x)=12(x^2-10x+25)$, whose roots are equal to $5$. We deduce that $a=b=c=d=5$. I used a consequence of Rolle's Theorem A slight imprecision. If $a,b,c,d$ are pairwise distinct, or if two of them are of equal value $v$ and the other two are either of distinct values different from $v$, or of equal value different from $v$, then indeed $P'(x)$ has three non-equal real roots. But if three of them are equal, and the fourth is of a different value, then $P'(x)$ has two equal roots, and a third of a different value, still however enough for $P''(x)$ to have two non-equal real roots - leading to the above contradiction. I don't see the "imprecision". Three non-equal roots means that the three roots are not all equal, as I see it.

Then I must beg your pardon. I read "three non-equal roots" as meaning "three distinct roots", seeing that "three roots, not all equal" would have been an equivalent, clearer (to me) way of expressing it ...

mavropnevma wrote: Then I must beg your pardon. No, it's my fault. There was a better way to point out that idea.

It's very easy to prove that $(a+b+c+d)^2\ge \frac{8}{3}(ab+ac+ad+bc+bd+cd)$: $(a+b+c+d)^2\ge \frac{8}{3}(ab+ac+ad+bc+bd+cd)\Leftrightarrow (a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2+(a-c)^2+(b-d)^2\ge 0$, which is obviously true. Since, $a + b + c + d = 20$ and $ab + ac + ad + bc + bd + cd = 150 \Rightarrow$ In the inequality I've proved, the equality holds. $\Rightarrow a=b=c=d$ $\Rightarrow a=b=c=d=5$.

In fact, there is a too simple solution: $$a^2+b^2+c^2+d^2=(a+b+c+d)^2-2(ab+ac+ad+bc+bd+cd)=20^2-2.150=100$$hence, $$(a-b)^2+(a-c)^2+(a-d)^2+(b-c)^2+(b-d)^2+(c-d)^2=3.100 -2.150=0$$which means that, $$a=b=c=d=5$$

We have \[a^2+b^2+c^2+d^2=(a+b+c+d)^2-2(ab+ac+ad+bc+bd+cd)=100.\]By Cauchy we have \[(a^2+b^2+c^2+d^2)(b^2+c^2+d^2+a^2)\geq (ab+bc+cd+ad)^2\]\[(a^2+b^2+c^2+d^2)(c^2+a^2+d^2+b^2)\geq (ac+ab+cd+db)^2\]\[(a^2+b^2+c^2+d^2)(c^2+d^2+b^2+a^2)\geq (ac+bd+bc+ad)^2\]and adding up these three inequalities and dividing by $2$ yields \[150=\frac{3}{2}(100)=\frac{3}{2}(a^2+b^2+c^2+d^2)\geq ab+ac+ad+bc+bd+cd\]and with equality iff $a=b=c=d=5$.

Note that \[ a^2+b^2+c^2+d^2 = 20^2 - 2 \cdot 150 = 100 \]but by Cauchy-Schwarz \[ 100 \cdot 4 = (a^2+b^2+c^2+d^2)(1+1+1+1) \ge (a+b+c+d)^2 = 400 \]so equality must hold meaning $a=b=c=d=5$, and this clearly works.

This was A2 on the shortlist.

Overcomplicated: The only possible $4$-tuple is $(a, b, c, d) = (5, 5, 5, 5)$, which clearly works. Now, we will show this is the only solution. It's clear that $$a^2 + b^2 + c^2 + d^2$$$$= (a+b+c+d)^2 - 2(ab + ac + ad + bc + bd + cd) = 100.$$Thus, we have $$\sum_{cyc} (a - 5)^2 = \left( \sum_{cyc} a^2 \right) - 10 \left( \sum_{cyc} a \right) + 4 \cdot 25$$$$= 100 - 10 \cdot 20 + 100 = 0$$so the trivial inequality implies $a = b = c = d = 5$ is the only possible solution. $\blacksquare$ Remark: We actually have $$\sum_{cyc} (a - k)^2 = (10 - 2k)^2$$for all real $k$. This result can be conjectured by plugging in small positive integers for $k$.

$$(a+b+c+d)^2=(20)^2$$$$\implies$$$$a^2+b^2+c^2+d^2=100$$$$\text{CS}$$$$4(a^2+b^2+c^2+d^2)\geq (a+b+c+d)^2$$$$\implies$$$$400\geq 400$$$$\implies$$$$a=b=c=d=5$$

Let S=sum of permutations of ab from a to d = 150. Then $400=(a+b+c+d)^2=a^2+b^2+c^2+d^2+2S$, or $a^2+b^2+c^2+d^2=400-2*150=100$. By AM-QM we have that $ \frac {a+b+c+d}{4} \le \sqrt {\frac {a^2 + b^2 + c^2 + d^2}{4}}$, with equality if and only if $ a=b=c=d$. But these expressions are equal, so equality holds at $a=b=c=d=5$.

uglysolutions wrote: Let $ S = ab + ac + ad + bc + bd + cd = 150$. We know that $ (a+b+c+d)^2 = a^2 + b^2 + c^2 + d^2 + 2S$, so $ a^2 + b^2 + c^2 + d^2 = 400 - 300 = 100$. So we have to find all real numbers $ a, b, c, d$ such that (1) $ a + b + c + d = 20$, and (2) $ a^2 + b^2 + c^2 + d^2 = 100$. By AM-QM inequality we have that $ \frac {a+b+c+d}{4} \le \sqrt {\frac {a^2 + b^2 + c^2 + d^2}{4}}$, with equality if and only if $ a=b=c=d$. But these expressions are in fact equal, as they both equal 5. So the only solution is $ a=b=c=d=5$