Inequalities: Star problems

Expansion gives

If $k>2$, division by the positive factor $k-2$ gives $x\leq k+2$.

If $k<2$, division reverses the sign and gives $x\geq k+2$.

At $k=2$ the original inequality is $0\leq0$, so every real x is a solution; this case cannot be recovered by division.

For part (b), $k=2$ fails because it includes both 4 and 1.

If $k>2$, the upper bound $k+2$ is greater than 4, so both numbers again satisfy the inequality.

Thus only $k<2$ can work.

In that case the condition is $x\geq k+2$.

The number 4 satisfies it automatically because $k+2<4$.

The number 1 fails exactly when $1<k+2$, or $k>-1$.

Combining the conditions gives $-1<k<2$.

At $k=-1$, the lower bound is exactly 1, so 1 is included and the requirement fails.

At $k=2$, the zero coefficient makes every real number a solution.

The two endpoints are excluded for different reasons.

Every integer, including a negative integer, can be written uniquely as $n=3q+r$, where $q$ is an integer and $r$ is one of $0,1,2$.

We inspect the interval in each case.

If $r=0$, the interval is $q<x<q+8/3$, so its integers are $q+1,q+2$.

If $r=1$, it is $q+1/3<x<q+3$, again allowing only $q+1,q+2$ because the upper endpoint is excluded.

If $r=2$, it is $q+2/3<x<q+10/3$, allowing $q+1,q+2,q+3$.

Therefore exactly three integers occur precisely when $n=3q+2$.

The remainder analysis exhausts all integers n and does not assume q is positive.

For these values, the sum of the three allowed integers is

Setting it equal to 30 gives $q=8$, hence $n=26$.

Directly, the interval is $26/3<x<34/3$, whose integers are indeed 9, 10, and 11.

Let $u=b-a\geq0$ and $v=c-b\geq0$, so $a=b-u$ and $c=b+v$.

Substitution cancels the terms containing b and gives

If $1\leq k\leq2$, both coefficients on the right are nonpositive.

Since u and v are nonnegative, the expression is at most zero for every ordered triple.

This argument allows negative a, b, or c because only their gaps matter.

If $k<1$, take

The expression is $1-k>0$, disproving the claim.

If $k>2$, take

The expression is $k-2>0$, again disproving it.

These counterexamples establish that no parameter outside the stated interval works.

For $1<k<2$, both gap coefficients are strictly negative, so equality requires $u=v=0$, or $a=b=c$.

At $k=1$, the expression is $-v$, giving equality exactly when $b=c$; a may be any smaller or equal number.

At $k=2$, it is $-u$, giving equality exactly when $a=b$; c may be any larger or equal number.

A solution must have $2x+3<5x-3$, so $x>2$.

For integer x this means $x\geq3$.

If $t\leq15$, the strict inequality $2x+3<t$ also forces $x<6$, leaving only 3, 4, and 5.

Their respective parameter intervals are $(9,12]$, $(11,17]$, and $(13,22]$.

For $t\leq15$, exactly two of these intervals overlap precisely on $(11,12]$ and $(13,15]$.

The strict left endpoints come from $2x+3<t$; the closed right endpoints come from $t\leq5x-3$.

In particular, $t=13$ allows only $x=4$, while $t=15$ allows $x=4,5$.

If $15<t<17$, the three integers $4,5,6$ all work.

For $t\geq17$, the allowed x interval is $[(t+3)/5,(t-3)/2)$, of length

Every interval that includes its left endpoint, excludes its right endpoint, and has length at least 3 contains at least three integers: take the first integer at or above the left endpoint and the next two.

The third is strictly less than the left endpoint plus 3.

Hence no $t>15$ yields exactly two solutions.

Together with the complete small-t analysis, this proves the two stated intervals are the entire answer.

Let G be the guaranteed score.

In particular, $G\leq x+2y$ and

Taking three copies of the first bound and two of the second gives

Therefore $G\leq36$ for every allowed allocation.

At $(x,y,z)=(0,18,12)$, the three scores are 36, 72, and 36, so the bound is attained.

If G equals 36, equality must hold throughout the displayed bound.

Thus $x=0$, and both positively weighted scores must equal G.

The equations $2y=36$ and $3z=36$ then force $y=18$ and $z=12$.

This proves uniqueness.

Making all three scores equal instead forces $x=2y=3z$.

Together with $x+y+z=30$, this gives $x=180/11$, $y=90/11$, and $z=60/11$, whose common score is $360/11<36$.

The second test need not be a limiting test at the optimum.

Spending resources to lower its excess score can improve the two scores that actually determine the guarantee.

Subtracting $3(x+y+z)=36$ from the second equation gives $-x+2z=14$, hence $x=2z-14$.

The sum equation then gives $y=26-3z$.

Positivity of x requires $z>7$.

The condition $x<y$ gives $2z-14<26-3z$, so $z<8$.

Finally, $y<z$ gives $z>13/2$, which is already guaranteed by $z>7$.

Therefore $7<z<8$.

Writing $z=7+s$ yields the displayed family with $0<s<1$.

Every such s satisfies all original equations and strict inequalities, so the description is both necessary and sufficient.

Now $x=2s$ runs through the open interval $0<x<2$.

Thus 2 is an upper bound, but it is never attained because $s=1$ would give $x=y=2$, violating the strict order.

There is no greatest x: from any $0<s<1$, replacing s by $(s+1)/2$ gives another allowed parameter and a strictly larger x.

Moreover, if $0<M<2$, choosing $s=(M+2)/4$ gives $x=(M+2)/2>M$; if $M\leq0$, every allowed positive x exceeds M.

Hence no number below 2 is an upper bound.

If $0\leq p<1$, then

Equality requires $x=0$ and $x+y=10$, giving $(0,10)$.

At $p=1$, every feasible point on $x+y=10$ is optimal.

Substitution into the other constraint gives $x\leq4$, so that full optimal edge is $0\leq x\leq4$, $y=10-x$.

If $1<p<2$, both coefficients in the identity

are positive.

Hence the value is at most

Equality requires both constraints to be tight, whose unique intersection is $(4,6)$.

At $p=2$, the second constraint directly gives maximum 14.

Points on $2x+y=14$ satisfy the first constraint exactly when $x\geq4$; nonnegativity gives $x\leq7$.

This yields the stated optimal edge.

Finally, if $p>2$, then

Equality requires $y=0$ and $2x+y=14$, giving $(7,0)$.

Each claimed point or segment is feasible, and the five cases cover every nonnegative p.

The weighted sum of the constraints gives

Equality would require all three pair sums to be tight.

But adding those three equalities would give $2(x+y+z)=49$, impossible for integers.

To measure the unavoidable loss, define nonnegative integer slacks $a=13-x-y$, $b=17-y-z$, and $c=19-z-x$.

Their sum is $49-2(x+y+z)$, an odd integer.

In particular, the slacks cannot all vanish.

The objective is exactly

Since at least one slack is a positive integer and all its weights are positive integers,

Thus the objective is at most 101.

Equality in this bound forces $a=1$ and $b=c=0$, because any positive b or c would cost at least 3 or 2.

The resulting equations are $x+y=12$, $y+z=17$, and $z+x=19$.

They give $(x,y,z)=(7,5,12)$, which is nonnegative, satisfies every original constraint, and has objective

The forced slack values and the unique system solution prove that no other triple attains the maximum.

Let R be the spread.

Every reading is at most the largest reading and at least the smallest, so $A-B\leq R$, $A-D\leq R$, and $C-B\leq R$.

Their expressions are $x-y$, $x+2y-12$, and $30-2x-3y$.

To choose useful weights u, v, w, require the variable terms in $u(A-B)+v(A-D)+w(C-B)$ to cancel.

This gives $u+v-2w=0$ and

Taking $w=3$ yields $u=1,v=5$, all positive.

With these weights,

Therefore $30\leq(1+5+3)R=9R$, giving the universal lower bound $R\geq10/3$.

This argument does not assume in advance which reading is largest or smallest.

To attain equality, all three positively weighted differences must equal R.

The equations $A-B=A-D$ and $A-B=C-B$ give $B=D$ and $A=C$.

Written in x and y, these are $3y=12$ and $3x+2y=30$.

Thus $y=4$ and $x=22/3$, both nonnegative.

At this pair, $A=C=56/3$ and $B=D=46/3$, so the spread is exactly $10/3$.

Hence the bound is attained.

Any optimal pair must satisfy the same equality conditions, whose solution is unique, proving the complete answer.