Given $a, b, c, d\in\mathbb{R^+}$ such that $abcd = 1$. Prove that
$a^ab^bc^cd^d\geq\frac{1}{16}(ab+c+d+1)^2$
I've tried,
$a^ab^bc^cd^d=(abcd)^ab^{b-a}c^{c-a}d^{d-a}=b^{b-a}c^{c-a}d^{d-a}\geq\frac{1}{16}(ab+c+d+1)^2$
i'm stucked
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Kuhaku
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3Where do you get this problem from? a homework ? or a test question ? or from a book ? if so what is the name of it? – Apr 26 '21 at 00:43
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This is a question taken from a free contest from Indonesia, called KTOM, the question is the last question. this the link https://drive.google.com/file/d/1xaYnrn4JiSnUyd8g6mOlPVGcM7jhp9VY/view?usp=sharing – Kuhaku Apr 26 '21 at 00:52
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1You might want to post this seemingly difficult question in MO ( mathoverflow ) where your chance of getting a good answer is much higher. Go there now. – Apr 26 '21 at 03:24
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4@NoTaNeViLsInDiSgUiSe: MO is about “questions related to current research in mathematics.” I am not sure if this question is on-topic on that site. – Martin R Apr 26 '21 at 07:38
1 Answers
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Since $x \mapsto x\ln x$ is convex on $x > 0$, and $a + b + c + d \ge 4\sqrt[4]{abcd} = 4$, we have \begin{align*} a\ln a + b\ln b + c\ln c + d\ln d &\ge 4 \cdot \frac{a + b + c + d}{4} \ln \frac{a + b + c + d}{4}\\ &\ge 4 \ln \frac{a + b + c + d}{4}. \end{align*} Also, we have $$ab + c + d + 1 \le \frac{1}{4}(a + b)^2 + c + d + 1.$$ It suffices to prove that $$\left(\frac{a + b + c + d}{4}\right)^4 \ge \frac{1}{16}\left(\frac{1}{4}(a + b)^2 + c + d + 1 \right)^2$$ or $$(a + b + c + d)^2 \ge (a + b)^2 + 4(c + d) + 4$$ or $$(c + d - 2)^2 + 2(a + b)(c + d) - 8 \ge 0$$ which is true since $2(a + b)(c + d) \ge 2\cdot 2\sqrt{ab} \cdot 2\sqrt{cd} = 8$.
We are done.
River Li
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@CalvinLin $f(x) = x \ln x$ is convex on $x > 0$ ($f''(x) = \frac{1}{x} > 0$). So, $\frac{f(a) + f(b) + f(c) + f(d)}{4} \ge f(\frac{a + b + c + d}{4})$. – River Li Apr 27 '21 at 00:30
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