$$a+b+c+d\ge4\sqrt[4]{abcd}=4\sqrt[20]{a^{5}b^{5}c^{5}d^{5}}\ge4\sqrt[20]{a^{2}b^{4}c^{6}d^{8}}=4\sqrt[10]{ab^{2}c^{3}d^{4}}=4.$$

Who is Janson Note that, using MA-MG: (1)$a+2b+3c+4d \ge 10\sqrt[10]{ab^2c^3d^4}=10\sqrt[10]{1}=10$. (2)$4a+3b+2c+d \ge 10\sqrt[10]{a^4b^3c^2d}=10\sqrt[10]{a(a)^3b^2(b)c^2d} \ge 10\sqrt[10]{a(d)^3b^2(c)c^2d}$ (because $a\ge d$ and $b\ge c$) $\Rightarrow 4a+3b+2c+d \ge 10\sqrt[10]{a(d)^3b^2(c)c^2d} =10\sqrt[10]{ab^2c^3d^4}=10$. (3)Adding (1) and (2): $5a+5b+5c+5d=5(a+b+c+d) \ge 10 + 10=20 \Rightarrow a+b+c+d \ge 4$.

Let $a_1 \geq a_2 \geq \cdots\geq a_n>0 $ and $a_1a_2^2\cdots a_n^n = 1$$(n\geq 3)$. Prove that $$a_1+ a_2+ \cdots +a_n\geq n$$

sqing wrote: Let $a_1 \geq a_2 \geq \cdots\geq a_n>0 $ and $a_1a_2^2\cdots a_n^n = 1$$(n\geq 3)$. Prove that $$a_1+ a_2+ \cdots +a_n\geq n$$ I'm suspecting AM-GM inequality

AbhayAttarde01 wrote: sqing wrote: Let $a_1 \geq a_2 \geq \cdots\geq a_n>0 $ and $a_1a_2^2\cdots a_n^n = 1$$(n\geq 3)$. Prove that $$a_1+ a_2+ \cdots +a_n\geq n$$ I'm suspecting AM-GM inequality Yeah same: Applying AM-GM on $a_1+2a_2+3a_3\dots+na_n$, we get that $\frac{a_1+2a_2+3a_3\dots+na_n}{ \frac{n(n+1)}{2}}\ge 1$. Because $a_1 \geq a_2 \geq \cdots\geq a_n$, we get that $a_n\ge 1$. Therefore, $a_1+\dots+a_n\ge n$.

EaZ_Shadow wrote: AbhayAttarde01 wrote: sqing wrote: Let $a_1 \geq a_2 \geq \cdots\geq a_n>0 $ and $a_1a_2^2\cdots a_n^n = 1$$(n\geq 3)$. Prove that $$a_1+ a_2+ \cdots +a_n\geq n$$ I'm suspecting AM-GM inequality Yeah same: Applying AM-GM on $a_1+2a_2+3a_3\dots+na_n$, we get that $\frac{a_1+2a_2+3a_3\dots+na_n}{ \frac{n(n+1)}{2}}\ge 1$. Because $a_1 \geq a_2 \geq \cdots\geq a_n$, we get that $a_n\ge 1$. Therefore, $a_1+\dots+a_n\ge n$. the top expression is not true consider if $a_i=10^{-143469420}$ for $i \neq 1$ and if $a_1$ is close to $\frac{n(n+1)}{2}$. then top expression still holds suspicious but to fix: there's a reason why we CANT make my assumption (think of original conditions) fixed thingy:

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