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Monday, November 30, 2009

Solution: Integer Inequality

Original problem: http://dharmath.thehendrata.com/2009/11/27/integer-inequality/

For any positive real number $t$, prove that there are integers $a,b,c,d$ such that

$! a^3b+b^3c+c^3d < tabcd$

Solution


Putting $a=b=c=d=1$, we find that the answer is trivial if $t > 3$. Now suppose $t \leq 3$.

Let $x=a/d, y=b/d, z=c/d$ be positive rational numbers, so our inequality becomes:

$! x^3y + y^3z +z^3 < txyz$

Let $f(x,y,z) = \frac{x^3y + y^3z +z^3}{xyz}$. Now, AM-GM asserts that $x^3y + y^3z + z^3 \geq 3xyz\sqrt[3]{yz}$, with equality happens if and only if $x^3y = y^3z =z^3$. In other words, when the equality conditions are satisfied, then $f(x,y,z) = 3\sqrt[3]{yz}$.

Equality conditions require that $z = y^{3/2}$ and $x = (y^2z)^{1/3} = y^{7/6}$. So, $f(y^{7/6}, y, y^{3/2}) = 3y^{5/6}$

If we want to find $y$ such that $3y^{5/6} < t$ then $y < (t/3)^{6/5}$. Which means we want to find $y$ such that $1/y > (3/t)^{6/5}$.

Now, let $N$ be an integer large enough such that $N^6 > (3/t)^{6/5}$. We can set $(x,y,z) = (1/N^7,1/N^6,1/N^9)$ which can be achieved by setting $(a,b,c,d) = (N^2, N^3, 1, N^9)$.

One can check that the assignment of $a,b,c,d$ does satisfy the condition of the problem:

$LHS = N^9 + N^9 + N^9 = 3N^9$

$RHS = t N^{14}$

$LHS < RHS \iff 3 < tN^5 \iff N^6 > (3/t)^{6/5}$, which is true by the way we chose $N$.

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