8
$\begingroup$

If $a,b,c,d>0$ satisfy the condition ${ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+{ d }^{ 2 }=1$, find the maximum value of $ab+ac+ad+bc+bd+3cd$.

I'm not progress in this inequality problem. Please help.

Thank you.

$\endgroup$
  • 4
    $\begingroup$ You could try langrange multipliers $\endgroup$ – XPenguen Dec 22 '15 at 15:08
  • 1
    $\begingroup$ Easily, but I'm in search of an elementary solution suitable for contest math. $\endgroup$ – Swapnil Das Dec 22 '15 at 15:21
  • $\begingroup$ Use Couchy inequality in first equation. Then consider the second. $\endgroup$ – openspace Dec 22 '15 at 15:26
11
$\begingroup$

Hint: consider that the eigenvalues of the symmetric matrix $$ M = \begin{pmatrix} 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 3 \\ 1 & 1 & 3 & 0 \end{pmatrix} $$ are $-3,-1,2+\sqrt{5}$ and $2-\sqrt{5}$. In particular, $\left(\frac{\sqrt{5}-1}{2},\frac{\sqrt{5}-1}{2},1,1\right)^T$ in the only eigenvector of $M$ with positive coordinates, and it is associated with the largest eigenvalue $2+\sqrt{5}$. It follows that:

$$ \max_{\substack{a^2+b^2+c^2+d^2=1 \\ a,b,c,d>0}} \frac{1}{2}(a\, b\, c\, d)\, M\, (a\, b\, c\, d)^T = \frac{2+\sqrt{5}}{2} = \color{red}{1+\frac{\sqrt{5}}{2}}. $$

Such a maximum is attained by $(a,b,c,d)=(x,x,y,y)$ with $x=\frac{1}{2}\sqrt{1-\frac{1}{\sqrt{5}}}$ and $y=\frac{1}{2}\sqrt{1+\frac{1}{\sqrt{5}}}$.

$\endgroup$
2
$\begingroup$

Let $ab+ac+ad+bc+bd+3cd=k$. Hence, $k>0$ and $ab+ac+ad+bc+bd+3cd=k(a^2+b^2+c^2+d^2)$ or $ka^2-(b+c+d)a+k(b^2+c^2+d^2)-bc-bd-3cd=0$. Hence, $(b+c+d)^2-4k(k(b^2+c^2+d^2)-bc-bd-3cd)\geq0$ or $(4k^2-1)b^2-2(2k+1)(c+d)b+(4k^2-1)(c^2+d^2)-2(6k+1)cd\leq0$. If $0<k\leq\frac{1}{2}$ so the last inequality is obviously true. Let $k>\frac{1}{2}$. Hence, $(2k+1)^2(c+d)^2-(4k^2-1)\left((4k^2-1)(c^2+d^2)-2(6k+1)cd\right)\geq0$ or $(2k+1)(c+d)^2-(2k-1)\left((4k^2-1)(c^2+d^2)-2(6k+1)cd\right)\geq0$ or $(2k^2-k-1)c^2-(6k-1)cd+(2k^2-k-1)d^2\leq0$. If $\frac{1}{2}<k\leq1$ so the last inequality is obviously true. Let $k>1$. Hence, $(6k-1)^2-4(2k^2-k-1)^2\geq0$, which gives $k\leq1+\frac{\sqrt5}{2}$. Easy to see that for $k=1+\frac{\sqrt5}{2}$ the equality indeed occurs. Id est, the answer is $1+\frac{\sqrt5}{2}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.