- #1
lfdahl
Gold Member
MHB
- 749
- 0
Suppose, the four real numbers $a,b,c$ and $d$ obey the inequality:$(a+b+c+d)^2 \ge K b c$, when $0 \le a \le b \le c \le d$.Find the largest possible value of $K$.
my solution:lfdahl said:Suppose, the four real numbers $a,b,c$ and $d$ obey the inequality:$(a+b+c+d)^2 \ge K b c$, when $0 \le a \le b \le c \le d$.Find the largest possible value of $K$.
I am interpreting the question to mean that we are looking for the largest value of $K$ such that $(a+b+c+d)^2 \geqslant K b c$ for all real numbers $a,b,c,d$ satisfying $0 \leqslant a \leqslant b \leqslant c \leqslant d$.lfdahl said:Suppose, the four real numbers $a,b,c$ and $d$ obey the inequality:$(a+b+c+d)^2 \ge K b c$, when $0 \le a \le b \le c \le d$.Find the largest possible value of $K$.
Thankyou very much, Opalg for another amazingly sharp and clear - and of course correct - answer! I also want to say, that I´m sorry, that I did not express my problem clear enough.Opalg said:I am interpreting the question to mean that we are looking for the largest value of $K$ such that $(a+b+c+d)^2 \geqslant K b c$ for all real numbers $a,b,c,d$ satisfying $0 \leqslant a \leqslant b \leqslant c \leqslant d$.
[sp]The numbers $a=0$, $b=c=d=1$ satisfy the given conditions. With those values, the inequality becomes $9\geqslant K$. On the other hand, the inequality $(a+b+c+d)^2 \geqslant 9 b c$ holds for all $a,b,c,d$ satisfying the given conditions.
To see that, notice first that the left side of the inequality becomes smaller if we put $a=0$ and $d=c$. So it will be sufficient to show that $(b+2c)^2 \geqslant 9bc$ whenever $0\leqslant b\leqslant c$. But $$(b+2c)^2 - 9bc = 4c^2 - 5bc + b^2 = (c-b)(4c-b),$$ and that is clearly $\geqslant 0$ whenever $c\geqslant b\geqslant 0.$
Therefore the largest possible value for $K$ is $9$.[/sp]
Albert said:my solution:
$K\leq \dfrac {(a+b+c+d)^2}{bc}$
$for \,\, a\leq b\leq c\leq d$
$if \,\, a,b,c\rightarrow 0$
$k \,\,depends \,\, on \,\, d\,\, only$
$so \,\, max(K)=\infty $
lfdahl said:Thankyou very much, Opalg for another amazingly sharp and clear - and of course correct - answer! I also want to say, that I´m sorry, that I did not express my problem clear enough.
- - - Updated - - -
Thankyou, Albert, for your participation. I think, the best, I can do, is to refer to Opalgs answer, and my short reply.
Opalg said:I am interpreting the question to mean that we are looking for the largest value of $K$ such that $(a+b+c+d)^2 \geqslant K b c$ for all real numbers $a,b,c,d$ satisfying $0 \leqslant a \leqslant b \leqslant c \leqslant d$.
[sp]The numbers $a=0$, $b=c=d=1$ satisfy the given conditions. With those values, the inequality becomes $9\geqslant K$. On the other hand, the inequality $(a+b+c+d)^2 \geqslant 9 b c$ holds for all $a,b,c,d$ satisfying the given conditions.
To see that, notice first that the left side of the inequality becomes smaller if we put $a=0$ and $d=c$. So it will be sufficient to show that $(b+2c)^2 \geqslant 9bc$ whenever $0\leqslant b\leqslant c$. But $$(b+2c)^2 - 9bc = 4c^2 - 5bc + b^2 = (c-b)(4c-b),$$ and that is clearly $\geqslant 0$ whenever $c\geqslant b\geqslant 0.$
Therefore the largest possible value for $K$ is $9$.[/sp]
Albert said:my explanation of this question is the following
first condition $0\leq a\leq b\leq c\leq d $ is given,and to satisfy the inequality,
$\dfrac {(a+b+c+d)^2}{bc}\geq {K}$,find $max(K)$
Is this interpretation reasonable ?
so we will have different answer to this question according to alien translation
ex :if a=b=c=d=1 then k=16, so k will be undefined and max(K) will be infinity
under this interpretation
I think this problem should be considered as an open question
Albert said:how about $a=0,b=c=0.1, d=1$
we have $(0+0.1+0.1+1)^2\geq K\times 0.1\times 0.1$
also satisfy the given inequility $1.44\geq 0.01K,and \,\, K\leq 144$
of course if we set $a=0,b=c=d=0.1,\,\,then\,\, K=9$
so in my opinion $9$ should be $min(K)$ instead of $max(K)$
for $a,b,c\in R $
and the inequility should be $(0\leq a<b\leq c\leq d)$
The purpose of finding the largest K is to determine the minimum value of K that satisfies the given inequality. This value of K represents the maximum amount that the product of bc can be less than the square of the sum of a, b, c, and d.
To solve for the largest K, you can expand the square of the sum and rearrange the terms to get Kbc on one side of the inequality. Then, divide both sides by bc to isolate K. The resulting value of K will be the largest possible value that satisfies the inequality.
Yes, the inequality can be solved algebraically by expanding the square of the sum and rearranging the terms to isolate K. However, the resulting expression may be complex and difficult to solve for K.
The largest K can be used in various real-life scenarios where the sum of a, b, c, and d represents a total amount and the product of bc represents a budget or limit. The value of K represents the maximum amount that the budget can be below the total without violating the given inequality.
No, the largest K in the inequality must be a positive number. This is because the square of the sum of any real numbers will always be positive, and the product of two negative numbers is also positive. Therefore, K must be positive in order for the inequality to hold true.