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This is somehow related to this problem but I don't have any idea about it.

Let $a$, $b$, $c$ and $d$ be positive reals such that $a+b+c+d=4$. Prove that: $$\frac{1}{a+3}+\frac{1}{b+3}+\frac{1}{c+3}+\frac{1}{d+3}\le \frac{1}{abcd}$$

Now I also tried to prove the $3$ variable version : $$\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\le \frac{1}{abc}$$ where $a,b,c>0$ with $a+b+c=3$. But I haven't been able to solve it too. Anyone can help?? Thanks a lot.

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$\sum$ will be used to denote the cyclic sum.

For the second inequality, let $t=ab+bc+ca$. Note that $t\le 3$, and also: $$3t=(a+b+c)(ab+bc+ca)\ge 9\sqrt[3]{(abc)(a^2b^2c^2)}=9abc$$ Now, by Cauchy-Schwarz: $$\sum\frac{2bc}{a+2}=2\sum\frac{(bc)^2}{abc+2bc}\ge \frac{2t^2}{3abc+2t}\ge \frac{2t}{3}$$ Note further that $\frac{2t}{3}-(t-1)=\frac{3-t}{3}\ge 0$ since $t\le 3$. Hence: $$\sum\frac{2bc}{a+2}\ge ab+bc+ca-1\implies \sum (bc-\frac{2bc}{a+2})\le 1\implies \sum\frac{1}{a+2}\le\frac{1}{abc}$$ as desired.

A similar trick works for the first inequality. Let $t=abc+bcd+cda+dab$ this time. Then by repeated applications of AM-GM, $$t=ab(c+d)+cd(a+b)\le (\frac{a+b}{2})^2(c+d)+(a+b)(\frac{c+d}{2})^2=\frac{(a+b+c+d)^2}{4}=4$$ Also, note that by AM-GM: $$4t=(a+b+c+d)(abc+bcd+cda+dab)\ge 16\sqrt[4]{(abcd)(a^3b^3c^3d^3)}=16abcd$$ Using these estimations: $$\sum \frac{3bcd}{a+3}=3\sum\frac{(bcd)^2}{abcd+3bcd}\ge \frac{3t^2}{4abcd+3t}\ge\frac{3t}{4}$$ And $\frac{3t}{4}-(t-1)=\frac{4-t}{4}\ge 0$, so that: $$\sum\frac{3bcd}{a+3}\ge abc+bcd+cda+dab-1\implies \sum (bcd-\frac{3bcd}{a+3})\le 1\implies \sum\frac{1}{a+3}\le \frac{1}{abcd}$$ as desired.

Note: Using this same method, the $n$-variable inequality can also be proven.

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We will prove a stronger inequality: If $a_1, a_2, \cdots, a_n$ are $k$ positive number such that $$\sum^{n}_{i=1}a_i = n$$ Then

$$\sum^{n}_{i=1} \dfrac{1}{a_i+n-1} \leq 1, \forall n \in \mathbb{N}$$

The above inequality is equivalent to

$$ \sum^{n}_{i=1} \Bigg( \dfrac{1}{n-1} - \dfrac{1}{a_i+n-1} \Bigg) \geq \dfrac{1}{n-1} $$ $$ \Leftrightarrow \sum^{n}_{i=1} \dfrac{a_i}{a_i + n - 1} \geq 1$$

Applying Schwarz inequality, we have $$ LHS = \sum^{n}_{i=1} \dfrac{(\sqrt{a_i})^2}{a_i+n-1} = P$$

It suffices to show that $$P \geq 1 \Leftrightarrow (\sum^{n}_{i=1}a_i)^2 \geq \sum^{n}_{i=1}a_i + n(n-1)$$ $$ \Leftrightarrow 2\sum_{1\leq i < j \leq n}\sqrt{a_ia_j} \geq n(n-1)$$

By applying AM-GM inequality for $\dfrac{n(n-1)}{2}$ numbers on the LHS, the conclusion follows.

P/S: This idea belongs to my Math teacher. He is very good at inequalities.

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  • $\begingroup$ If $\sum_{i=1}^na_i=n$, then in fact $$n^2\sum\frac{1}{a_i+n-1}=(\sum (a_i+n-1))(\sum\frac{1}{a_i+n-1})\ge n^2$$ by Cauchy-Schwarz, so $\sum\frac{1}{a_i+n-1}\ge 1$. $\endgroup$ – Apple Jul 23 '14 at 18:17
  • $\begingroup$ Wow that was much faster :) I checked my note and the condition in my problem is $a_1a_2\cdots a_n = n$. I think it's a good way to strengthen the inequality :) $\endgroup$ – primitiveroot Jul 23 '14 at 18:27
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Let $a+b+c+d=4u$ and $ab+ac+ad+bc+bd+cd=6v^2$.

Hence, by AM-GM $abcd\leq v^4$ and by C-S $$\frac{1}{abcd}-\sum_{cyc}\frac{1}{a+3}=\frac{1}{abcd}-\frac{4}{3}-\sum_{cyc}\left(\frac{1}{a+3}-\frac{1}{3}\right)=$$ $$=\frac{1}{abcd}-\frac{4}{3}+\sum_{cyc}\frac{a}{3(a+3)}\geq\frac{1}{abcd}-\frac{4}{3}+\frac{(a+b+c+d)^2}{3\sum\limits_{cyc}(a^2+3)}=$$ $$=\frac{1}{abcd}-\frac{4}{3}+\frac{16}{3\sum\limits_{cyc}(a^2+3)}=\frac{u^2}{abcd}-\frac{4}{3u^2}+\frac{16}{3(16u^2-12v^2+12u^2)}\geq$$ $$\geq\frac{u^2}{v^4}-\frac{4}{3u^2}+\frac{4}{3(7u^2-3v^2)}\geq0,$$ where the last inequality it's $$(u^2-v^2)(7u^4+4u^2v^2-4v^4)\geq0,$$ which is true because $u^2\geq v^2$.

Done!

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