pieater314159 wrote: Let $a$, $b$, $c$ be positive reals such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$. Show that $$a^abc+b^bca+c^cab\ge 27bc+27ca+27ab.$$ Proposed by Milan Haiman Solution. Since $a,b,c>0$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$, we get $\frac{3}{\sqrt[3]{abc}}\le1$ thereby $abc\ge27$, and so the Weighted AM-GM shows \begin{align*}&\frac{a^a}{a}+\frac{b^b}{b}+\frac{c^c}{c}\ge (a^a)^\frac{1}{a} (b^b)^\frac{1}{b} (c^c)^\frac{1}{c}=abc\ge27, \end{align*}which gives $a^abc+b^bca+c^cab\ge 27abc= 27bc+27ca+27ab$. As desired. $\blacksquare$

Weighted AM-GM $$\sum_{cyc}\frac{bc}{ab+bc+ca}\cdot a^a\ge \prod_{cyc} a^\frac{abc}{ab+bc+ca}=abc$$GM-HM $$abc\ge\left(\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\right)^3=27$$Equality iff $a=b=c=3$

ytChen wrote: pieater314159 wrote: Let $a$, $b$, $c$ be positive reals such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$. Show that $$a^abc+b^bca+c^cab\ge 27bc+27ca+27ab.$$ Proposed by Milan Haiman Solution. Since $a,b,c>0$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$, we get $\frac{3}{\sqrt[3]{abc}}\le1$ thereby $abc\ge27$, and so the Weighted AM-GM shows \begin{align*}&\frac{a^a}{a}+\frac{b^b}{b}+\frac{c^c}{c}\ge (a^a)^\frac{1}{a} (b^b)^\frac{1}{b} (c^c)^\frac{1}{c}=abc\ge27, \end{align*}which gives $a^abc+b^bca+c^cab\ge 27abc= 27bc+27ca+27ab$. As desired. $\blacksquare$ Truly amazing.

consider the function $f(x)=x^x$ Note that $f'(x)=x^x(1+lnx)$ which is defined and positive for all positive reals.Therefore the function is concave upwards. Note that the given statement is equivalent to $ab+bc+ca=abc$..............(1) By Jensen on $f(x)$, we get that $\frac{a^a}{a}+\frac{b^b}{b}+\frac{c^c}{c}\geq (\frac {a}{a}+\frac{b}{b}+\frac{c}{c})^{a+b+c}=3^{a+b+c}$....(2) Also observe that $a+b+c\geq 3$ because if their sum is less than 3 then one of them is less than 1 implying its reciprocal is greater than 1 and thus the sum of the reciprocals of the numbers is greater than 1 which is a contradiction. Thus $3^{a+b+c}\geq 3^3=27$........(3) From (1),(2) and (3) we get that $a^{a}bc+ab^{b}c+abc^{c}\geq 27(ab+bc+ca)$

Woww! I wish this had come on the actual test. Not because it is easy, and it indeed isn't, atleast for me, I took an hour to get here tbh. I wish this was on the actual test because I like solving such Inequalities. Anyways, here's the proof. RHS equals 27abc So basically what were left to prove is $a^{a-1}+b^{b-1}+c^{c-1}\geq 27$ Let $w_1=a^{-1}, w_2=b^{-1}, w_3=c^{-1}, a_1=a^a, a_2=b^b, a_3=c^c$ By weighted AM-GM, $\frac{a_1w_1+a_2w_2+a_3w_3}{w_1+w_2+w_3}\geq (a_1^{w_1}a_2^{w_2}+a_3^{w_3})^{\frac{1}{w_1+w_2+w_3}}$ Which gives $a^{a-1}+b^{b-1}+c^{c-1}\geq abc$. But by GM HM, $abc\geq 27$. Thus, proved

Since $f(x)=x^x$ is convex, applying weighted Jensen (with weights being $\frac{1}{a}$, etc.) gives $$\frac{a^a}{a}+\frac{b^b}{b}+\frac{c^c}{c}\geq \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\right)^\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=27$$Then using the condition, multiplying both sides by $abc=ab+ac+bc$ gives $a^abc+b^bca+c^cab\geq 27ab+27ac+27bc$ as desired.

$ My $ $ solution $ Lemma(Young) :$ \frac{1}{p}+\frac{1}{q}+\frac{1}{r}=1, $ $Then $ $ \frac{a^p}{p}+\frac{b^q}{q}+\frac{c^r}{r}\geq abc $ Proof: $ Karamata $ $ inequality $ $ p=a,q=b,r=c $ $ \frac{a^a}{a}+\frac{b^b}{b}+\frac{c^c}{c}\geq abc\geq27 $

Claim: $abc\ge27.$ Proof. Observe that since $a,b,c$ are positive, then by GM-HM, $$\sqrt[3]{abc}\ge\frac{3}{\tfrac{1}{a}+\tfrac{1}{b}+\tfrac{1}{c}}.$$Substituting the given condition gives the desired. $\blacksquare$ Claim: $a^{a-1}+b^{b-1}+c^{c-1}\ge abc.$ Proof. Note that since $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$, then by Weighted AM-GM $$a^{a-1}+b^{b-1}+c^{c-1}=\left(\frac{1}{a}\right)a^a +\left(\frac{1}{b}\right) b^b+\left(\frac{1}{c}\right) c^c\ge a^{a(1/a)}b^{b(1/b)}c^{c(1/c)}=abc.$$ Hence, $a^{a-1}+b^{b-1}+c^{c-1}\ge27,$ and multiplying by $abc$ and using the given condition yields $$a^abc+b^bac+c^cab\ge 27abc=27(ab+bc+ca),$$as desired. $\square$

pieater314159 wrote: Let $a$, $b$, $c$ be positive reals such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$. Show that $$a^abc+b^bca+c^cab\ge 27bc+27ca+27ab.$$ Proposed by Milan Haiman $a^abc+b^bca+c^cab=\frac 1a (a^{a+1}bc)+\frac 1b (ab^{b+1}c)+\frac 1c (abc^{c+1}) $ Now apply Weighted AM-GM $\frac 1a (a^{a+1}bc)+\frac 1b (ab^{b+1}c)+\frac 1c (abc^{c+1}) \geq (\frac 1a+\frac 1b+\frac 1c)[(a^{1+\frac 1a}b^{\frac 1a}c^{\frac 1a})(a^{\frac 1b}b^{1+\frac 1b}c^{\frac 1b})(a^{\frac 1c}b^{\frac 1c}c^{1+\frac 1c})]^{\frac 1{\frac 1a+\frac 1b+\frac 1c}}=(abc)^2$ And now by AM-HM on one $abc$ we get $(abc)^2=abc(abc)\geq abc (\frac{3}{\frac 1a+\frac 1b+\frac 1c})^3=27abc=27abc(\frac 1a+\frac 1b+\frac 1c)=27bc+27ca+27ab$ Hence proved . Equality iff $a=b=c=3$

Notice that it's equivalent to $$\frac{1}{a}a^a+\frac{1}{b}b^b+\frac{1}{c}c^c\geq 27(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=27$$By Weighted AM-GM we have $$\frac{1}{a}a^a+\frac{1}{b}b^b+\frac{1}{c}c^c\geq abc$$Now it suffices to show that $abc \geq 27$. Indeed, by GM-HM $\sqrt[3]{abc}\geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=3 \implies abc \geq 27$. $\square$

A really nice and tricky ineq. ELMO SL 2019 A1 wrote: Let $a$, $b$, $c$ be positive reals such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$. Show that $$a^abc+b^bca+c^cab\ge 27bc+27ca+27ab.$$ Solution. By dividing by $abc$ we know that it is enough to show that:$\frac{1}{a}a^a+\frac{1}{b}b^b+\frac{1}{c}c^c\geq 27(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=27$. By Weighted AM-GM and GM-HM inequalities we have that: $$\frac{1}{a}a^a+\frac{1}{b}b^b+\frac{1}{c}c^c\geq abc=(\sqrt[3]{abc})^3\stackrel{GM-HM}{\geq}(\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}})^3=3^3=27.\blacksquare$$

Oops. Let $a=\frac{1}{x},b=\frac{1}{y},c=\frac{1}{z}$. Then $x+y+z=1$ and the inequality becomes: $$\sum_{cyc} \frac{1}{yz\sqrt[x]{x}} \geq 27\sum_{cyc} \frac{1}{xy}.$$Then multiplying by $xyz$, this is equivalent to: $$x^{x-1/x}+y^{y-1/y}+z^{z-1/z} \geq 27(x+y+z)=27.$$Note that the function $f(x)=x^{1-1/x}$ is convex on $(0,1)$ (I'll skip the details here, but the idea is to take the log of both sides, differentiate, use a few rules, and get the original derivative), hence by Jensen's: $$x^{1-1/x}+y^{1-1/y}+z^{1-1/z} \geq 3\left(\frac{1}{3}\right)^{1-3}=27. ~\blacksquare$$

pieater314159 wrote: Let $a$, $b$, $c$ be positive reals such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$. Show that $$a^abc+b^bca+c^cab\ge 27bc+27ca+27ab.$$ Proposed by Milan Haiman Note that: $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1 \implies bc+ca+ab=abc$$Hence we will re-write the problem as show that: $$a^abc+b^bca+c^cab\ge 27(bc+ca+ab)=27abc \implies a^{a-1}+b^{b-1}+c^{c-1} \ge 27$$The last one can be proved by Jensen's : Note that $(a, b, c) > 1$ , now let $f(x)=x^x$ then: $$\frac{d^2}{dx^2} f(x) = x^{x-1}+(\ln(x)+1)^2 \cdot x^x$$This means that $f(x)$ is convex in $\mathbb R^+$. Now we will use Jensen's inequality with: $$\left \{ f(x)=x^x \mid (a_1, a_2, a_3)=\left(\frac{1}{a}, \frac{1}{b}, \frac{1}{c} \right) \mid (x_1, x_2, x_3) = (a,b,c) \mid [a, b]=\mathbb R^+ \right \}$$$$\frac{f(a)}{a}+\frac{f(b)}{b}+\frac{f(c)}{c} \ge f(1+1+1) \implies a^{a-1}+b^{b-1}+c^{c-1} \ge 27$$Equality gives when $a=b=c=3$ Thus we are done . (i'm happy becuase this is my first inequality solved by Jensen's )

A slightly different solution

By Weighted AM-GM, $\sum_{\text{cyc}}\frac{a^a}a\ge abc$, so it suffices to show that $abc\ge27$. This is true since $abc=ab+bc+ca\ge3(abc)^{2/3}$.

The inequality is equivalent to $$\sum_{cyc} a^{a-1}\geq 27.$$Define $f(x)=x^{x-1}$ for $x>1$. Then note that $f$ is increasing on $x> 1$ and convex on $x> 1$. Hence, we apply Jensen's as $a,b,c>1$, $$\sum_{cyc} a^{a-1}\geq 3 \left(\frac{a+b+c}{3}\right)^{\frac{a+b+c}{3}-1}\geq 3 \left(\frac{9}{3}\right)^{\frac{9}{3}-1}=27,$$where we used $a+b+c\geq \dfrac{9}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=9$.

I liked this inequality. This inequality is equivalent to show $$\sum_{cyc}a^{a-1}=27$$and we know that $\sum_{cyc}\frac{1}{a}=1$ By weighted AM-GM,we have $$\sum_{cyc}\frac{1}{a}(a^a) \geq abc\stackrel{GM-HM}{\geq}27$$

Power mean inequality

Imagine Jensen with a neat function. Couldn't be me

The inequality is equivalent to $$a^abc+b^bca+c^cab\ge 27bc+27ca+27ab=27(bc+ca+ab)=27abc \rightarrow a^{a-1}+b^{b-1}+c^{c-1} \geq 27.$$Let $f(x)=x^{x-1}$. With a quick calculation, we get that $f$ is convex, so by Jensen's we have $$f(a)+f(b)+f(c) \geq 3f\left(\frac{a+b+c}{3}\right),$$so it suffices to show $a+b+c \geq 9$. But this is is just $AM-HM$ on $a,b,c$: $$\frac{a+b+c}{3} \geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}} \rightarrow a+b+c \geq 9$$as required. $\blacksquare$

Let $x=\frac1a$, $y=\frac1b$, $z=\frac1c$. Rearrange the wanted inequality to \[x^{1-1/x}+y^{1-1/y}+z^{1-1/z}\ge 27.\]

Therefore, $f(x)+f(y)+f(z)\ge 3f\left(\frac13\right)=27$. QED.

Note that $abc=ab+ac+bc$, so this becomes $$a^abc+ab^bc+abc^c\geq 27abc,$$or $$a^{a-1}+b^{b-1}+c^{c-1}\geq 27.$$Let $x=1/a,y=1/b,z=1/c$, so that $x+y+z=1$ and this becomes $$x^{1-1/x}+y^{1-1/y}+z^{1-1/z}\geq 27.$$Let $f(x)=x^{1-1/x}$. By logarithmic differentiation $f$ is convex over $(0,1)$, so we have $$f(x)+f(y)+f(z)\geq 3f(1/3)=27,$$as desired.

Dividing through by $abc$, it suffices to show that \[ \sum_{\text{cyc}} a^{a-1} \ge 27. \]Write $(x, y, z)=(1/a, 1/b, 1/c)$. Note that $f(x)=x^{1-1/x}$ is convex over $(0, 1)$, which can be seen through logarithmic differentiation, so Jensen's implies \[ \sum_{\text{cyc}} a^{a-1} = \sum_{\text{cyc}} f(x) = 3 \sum_{\text{cyc}} \frac{1}{3} \cdot f(x) \ge 3 f(\sum_{\text{cyc}} \frac{1}{3} x) = 3f(1/3)=27, \]as desired.

Weighted AM-GM gives $$\frac{a^a\cdot bc + b^b \cdot ac + c^c\cdot ab}{ab+bc+ac}\geq \sqrt[ab+bc+ac]{a^{abc}b^{abc}c^{abc}}=\sqrt[abc]{a^{abc}b^{abc}c^{abc}}=abc.$$Thus it suffices to show that $abc\geq 27,$ which follows from the fact that $$\frac{1}{3}=\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3}\geq \sqrt[3]{\frac{1}{abc}}.$$

Notice that the inequality is equivalent to $\sum_{cyc}a^abc\ge27\sum_{cyc}ab=27abc\Longleftrightarrow\frac{1}{a}a^a+\frac{1}{b}b^b+\frac{1}{c}c^c\ge27$ Furthermore $\frac{1}{a}a^a+\frac{1}{b}b^b+\frac{1}{c}c^c\overset{\text{Weighted AM-GM}}{\ge}(a^a)^{\frac{1}{a}}(b^b)^{\frac{1}{b}}(c^c)^{\frac{1}{c}}=abc$ Moreover notice that $1=\sum_{cyc}\frac{1}{a}\ge\frac{3}{\sqrt[3]{abc}}\Longrightarrow abc\ge27$ Thus $\frac{1}{a}a^a+\frac{1}{b}b^b+\frac{1}{c}c^c\ge abc\ge27\Longrightarrow\sum_{cyc}a^abc\ge27abc=27\sum_{cyc}ab$ Therefore $\sum_{cyc}a^abc\ge27\sum_{cyc}ab$ as desired $\blacksquare$.

Rearrange to $$abc(a^{a - 1} + b^{b - 1} + c^{c - 1}) \ge 27(ab + bc + ca) \Longleftrightarrow a^{a - 1} + b^{b - 1} + c^{c - 1} \ge 27,$$but by Weighted AM-GM, we have $$\frac{a^a \cdot \frac{1}{a} + b^b \cdot \frac{1}{b} + c^c \cdot \frac{1}{c}}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \ge \sqrt[\frac{1}{a} + \frac{1}{b} + \frac{1}{c}]{abc},$$after which simplifying gives $a^{a - 1}b^{b - 1}c^{c - 1} \ge abc$, but we also have $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1 \ge \frac{3}{\sqrt[3]{abc}},$ so $abc \ge 27$, which gives the desired.

Observe that $a^{a-1} + b^{b-1} + c^{c-1}$ $\ge 3(a^{a-1} \cdot b^{b-1} \cdot c^{c-1})^{\frac{1}{3}}$ $\ge 3\left(\frac{a+b+c-3}{\frac{a-1}{a} + \frac{b-1}{b} + \frac{c-1}{c}}\right)^{\frac{a+b+c}{3}-1}$ $= 3\left(\frac{a+b+c-3}{2}\right)^{\frac{a+b+c}{3}-1}$ $\ge 3\left(\frac{a+b+c-3}{2}\right)^{2}$ (*) $\ge 3\left(\frac{\frac{9}{2}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\right)^{2} = 27$ Now multiplying with $abc$, we are done. Note: (*) comes from observing that $a+b+c-3 > 2$, as one of $a, b, c$ has to be more than or equal to $3$ and all of them are more than $1$, then applying AM-GM.

$\frac{a^abc+b^bac+c^cab}{ab+bc+ca} \geq abc$ we need to prove $abc \geq 27$ $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 3\frac{1}{(abc)^{\frac{1}{3}}}$ So, $abc \geq 27$

Claim: $x^x$ is convex over $\mathbb R^{+}$. Proof. First derivative is just $x^x(\ln x+1)$, while second derivative is $x^x(\ln x +1)+x^x(\ln x+1)\ln x + x^{x-1}=x^{x-1}(2x\ln x+x+x(\ln x)^2+1)>0$. $\blacksquare$ We can now apply Jensen's inequality. Note, \[\frac1aa^a+\frac1bb^b+\frac1cc^c\geq\left(\frac1a+\frac1b+\frac1c\right)\left(\frac{a+b+c}3\right)^{\frac{a+b+c}{3}}=\left(\frac{a+b+c}3\right)^{\frac{a+b+c}{3}}\]Note that since $\frac1a+\frac1b+\frac1c=1$, we have $a+b+c\geq9$ by AM-HM inequality. So, $\frac{a+b+c}{3}\geq1$, and we know that for $x\geq1$, $x^x$ is increasing, which means \[\left(\frac{a+b+c}3\right)^{\frac{a+b+c}{3}}\geq3^3=27\]Combining, we get \[\frac1aa^a+\frac1bb^b+\frac1cc^c\geq27\Longrightarrow abc\left(\frac1aa^a+\frac1bb^b+\frac1cc^c\right)\geq27(bc+ca+ab),\]where we have used $bc+ca+ab=abc$ because $\frac1a+\frac1b+\frac1c=1$. The proof is complete. $\blacksquare$

Divide the inequality by $abc$. We get $$\frac{1}{a}a^a+\frac{1}{b}b^b+\frac{1}{c}c^c \geq 27$$By the weighted AM-GM, $$\frac{1}{a}a^a+\frac{1}{b}b^b+\frac{1}{c}c^c \geq abc \geq (\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}})^3 =27$$

the $weighted$ $ power $ $ mean$ kills it basically

Note that by Young's Inequality for $3$ variables we have that $$\text{LHS} \geq (abc)^2,$$so it suffices to show that \begin{align*} (abc)^2 \geq 27(ab+bc+ca) &\iff abc \geq 27 \\ &\iff \frac{1}{3} \geq \frac{1}{\sqrt[3]{abc}} \\ \end{align*}which holds by AM-GM. QED

Cool Inequality, I pretty much headsolved this while walking to physics class. We start off with the following observation. Claim : $abc \ge 27$ Proof : Let $a+b+c=3u$ , $ab+bc+ca=3v^2$ and $abc=w^3$. Then, it is well known that $u\ge v \ge w$. Now, note that the condition that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$ rewrites to $abc=ab+bc+ca$ and thus, \[w^3 = 3v^2 \ge 3w^2\]So, $w\ge 3$ and thus, \[abc=w^3 \ge 27\]as desired. With this observation in hand, we can finish off the problem. Note that by Weighted AM-GM we have, \[a^{a-1}+b^{b-1}+c^{c-1} \ge \sqrt[\uproot{5} \frac{1}{a}+\frac{1}{b}+\frac{1}{c}]{abc}=abc\]Now, \[a^abc + b^bca + c^2ab \ge (abc)^2 \ge 27(abc)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=27bc + 27ca + 27ab\]with equality if and only if $a=b=c=3$, which proves the desired inequality.