MHB Inequality - find the largest K in (a+b+c+d)^2≥Kbc

[lfdahl]

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 $K$.

 

[Albert1]

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:

Spoiler

$K\leq \dfrac {(a+b+c+d)^2}{bc}$
$for \,\, a\leq b\leq c\leq d$
$if \,\, a,b,c\rightarrow 0$
$k \,\,depends \,\, on \,\, d\,\, only$
$so \,\, max(K)=\infty $

 

[Opalg]

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 $0 \leqslant a \leqslant b \leqslant c \leqslant d$.

[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]

 

[lfdahl]

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]

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.

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my solution:

Spoiler

$K\leq \dfrac {(a+b+c+d)^2}{bc}$
$for \,\, a\leq b\leq c\leq d$
$if \,\, a,b,c\rightarrow 0$
$k \,\,depends \,\, on \,\, d\,\, only$
$so \,\, max(K)=\infty $

Thankyou, Albert, for your participation. I think, the best, I can do, is to refer to Opalgs answer, and my short reply.

 

[Albert1]

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.

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Thankyou, Albert, for your participation. I think, the best, I can do, is to refer to Opalgs answer, and my short reply.

Spoiler

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

 

[Albert1]

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]

Spoiler

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 $(0\leq a<b\leq c\leq d)$

 

[lfdahl]

Spoiler

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

Hi, Albert

Spoiler

I think, the point is, that the $K$-value can only be determined, when we consider all possible values of $a,b,c$ and $d$.
For example: Letting $a=b=c=d=1$ does not imply, that $K = 16$. The case: $a = 0$, $b=c=d=1$ should also obey the
inequality. I guess, that´s why Opalg underlines "all" ($a, b, c, d$ values) in his interpretation of the problem. It is true, that if we stick to some few special cases, $K$ will be undefined, as you notice. But using only some cases to determine $K$ is a misinterpretation of the problem. I admit, that my way of formulating the problem is not clear enough, and I´m sorry for this.

 

[lfdahl]

Spoiler

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 $(0\leq a<b\leq c\leq d)$

Hello again, Albert!

Spoiler

It is indeed a reasonable consideration: whether we look for a maximum or a minimum value of $K$!
I´d say, we are looking for a maximum value, because:
Any value of $K$ below $K_{max}$ makes the inequality statement true, thus $K \in ]-\infty; K_{max}]$ implies, that any choice of $a,b,c,d$ (of course still with the restriction: $0 \le a \le b \le c \le d$) is accepted/valid. In other words: It is easy for us to make the inequality true, we can simply pick any suitable low $K$. But the opposite is not true: We cannot pick an arbitrary high $K$-value, simply because we have to consider all possible values of $a,b,c$ and $d$.