This inequality $a+b^2+c^3+d^4\ge \frac{1}{a}+\frac{1}{b^2}+\frac{1}{c^3}+\frac{1}{d^4}$

Question

let $0<a\le b\le c\le d$, and such $abcd=1$,show that $$a+b^2+c^3+d^4\ge \dfrac{1}{a}+\dfrac{1}{b^2}+\dfrac{1}{c^3}+\dfrac{1}{d^4}$$

it seems harder than This inequality $a+b^2+c^3\ge \frac{1}{a}+\frac{1}{b^2}+\frac{1}{c^3}$

Answer 1

To take account of the restriction $0 \le a \le b \le c \le d$ we define three variables $u,v,w$ which will each be restricted to $[1,\infty)$ and put $b=au^4,\ c=bv^4,\ d=cw^4.$ The reason for the fourth powers is to have simple representations for $a,b,c,d$ which make them also satisfy the restriction $abcd=1.$ This is an iff type condition which then leads to unique expressions for each of $a,b,c,d$ namely $$ a=\frac{1}{u^3v^2w} \\ b=\frac{u}{v^2w} \\ c=\frac{uv^2}{w} \\ d=uvw^2.$$

Now let $f(a,b,c,d)=a-1/a+b^2-1/b^2+c^3-1/c^3+d^4-1/d^4.$ the expression we wish to show is nonnegative. We substitute in $f$ the expressions above for $a,b,c,d$ in terms of $u,v,w$ and get an equivalent expression $g(u,v,w)$ in the new variables. The advantage so far is that there are now no restrictions on $u,v,w$ other than that each is in $[1,\infty)$

Using a CAS we obtain for $g(u,v,w)$ an expression having denominator $u^4v^8w^{12}$ and its numerator is $m(u,v,w)$ where $$m(u,v,w)=u^8v^{16}w^{24}+(v^4-w^4)[u^7v^{10}w^9+uv^2w^{11}] \\ +u^6v^4w^{10}-u^2v^{12}w^{14}-1.$$ If now it happens that $v \ge w$ the value of $m$ is easily seen to be nonnegative, since (recall $u,v,w \ge 1$) the first term dominates the variable subtracted one, and also clearly $u^6v^4w^{10}-1 \ge 0.$

The remaining case is that in which $v<w,$ and I could see no slick way to complete this case. However it is indeed provable by defining (after changing the variables for no good reason but my notes) the polynomial $h(x,y,t)=m(x,y,y+t).$ So here $t$ is measuring the extent by which the third of $u,v,w$ exceeds the second, to reflect we're looking at the remaining case $v<w.$ When this $h(x,y,t)$ is expanded into a polynomial in $t$ of degree 24, with coefficients each polynomials in $x$ and $y$, it is fairly easy to see these coefficients are each nonnegative, though admittedly tedious. For example the coefficient of $t^4$ is $$xy^{10}(10626x^7y^{26}-589x^6y^9+210x^5-1001xy^{12}-1035y^3).$$ Here the first term dominates the subtracted ones in terms of each of its exponents exceeding those of other terms, and also (rather clearly in this case) the coefficient contribution from the first term more than offsets that of the subtracted terms. It goes the same for all the other terms, I just don't know if anyone wants me to type it all in (I'd rather not).

I'd like to see a slicker proof at least of the case $v<w.$

[I deleted a proposed simplification since to show a polynomial has nonnegative derivatives involves showing all of its (the derivative's) coefficients nonnegative, which is basically done as above with no easy short-cut.]

Answer 2

it is trivial $d \ge 1$

there are two cases: $c\ge 1$ , or $c\le 1$

$c\ge 1$

let $x=\dfrac{1}{a},y=\dfrac{1}{b} \implies x\ge y\ge 1 ,xy=cd$

let $f(x)=x-\dfrac{1}{x}$, it is trivial $f(x)$ is mono increasing

function

now we need to prove :

$f(c^3)+f(d^4) \ge f(x)+f(y^2)$

1)if $ c^3< x \cap c^3<y^2 \implies yc^3 <xy=cd \iff d> c^2y>y \cap x^2c^3 <x^2y^2=c^2d^2 \iff d^2>cx^2 \iff d>x $

$f(d^4)=f(d^2)(d^2+\dfrac{1}{d^2}) \ge 2 f(d^2) > f(x^2)+f(y^2)>f(x) +f(y^2)$

$f(c^3) \ge 0$ and this part is true.

2)if $x>y^2 \cap y^2 <c^3<x \implies c^3d^3=x^3y^3 <xd^3 \iff d^3>y^3x^2 \iff d^4>x^2 $

$f(c^3)+f(d^4) >f(y^2)+f(x^2)>f(x)+f(y^2)$

3)if $x<y^2 \cap x<c^3<y^2 \implies x^3y^3=c^3d^3 <y^2d^3\iff d^3>x^2y \ge y^3 \iff d>y$

$f(c^3)>f(x), f(d^4)>f(y^4)>f(y^2)$ it is true also.

4)it is trivial true when $c^3>x,c^3>y^2$

when $c \le 1 ,z=\dfrac{1}{c} \implies x \ge y \ge z \ge 1, d=xyz \ge 1$

we need to prove : $ f(d^4) \ge f(x)+f(y^2)+f(z^3)$

$f(z^4) \ge f(z^3),\iff f((xyz)^4) \ge f(x)+f(y^2)+f(z^4) \\ \iff (xy)^4((xy)^4-1)v^2+x^2y^2(x^2+xy^2-x^3y^2-x^2y^4)v+(xy)^4-1 \ge 0 ,v=z^4$

$(xy)^4((xy)^4-1)>0,$ if $\Delta \le 0 $ then it is true.

$\Delta \le 0 \\ \iff (x^2+xy^2-x^3y^2-x^2y^4)^2 \le 4((xy)^4-1)^2 \\ \iff x^2y^4+x^3y^2-x^2-xy^2 \le 2((xy)^4-1) \\ \iff (2x^4-x^2)y^4-(x^3-x)y^2+x^2-2 \ge 0 $

$g(t=y^2)=(2x^4-x^2)t^2-(x^3-x)t+x^2-2$

note $\dfrac{(x^3-x)}{2(2x^4-x^2)}<1 \iff 4x^3 -x^2 -2x+1 \ge 0 \iff 2x^2(2x-1)+(x-1)^2 \ge 0$ is true,

$g_{min}=g(1)=2x^4-x^3+x-2=(x^2-1)(2x^2-x+2) \ge 0$

Answer 3

Here is an "almost complete" solution except for the last step, where I'm frankly a bit stuck. It also points in the direction of a general algorithm in solving these kind of $\sum_{i=1}^{n}x_i^i \ge \sum_{i=1}^{n}x_i^{-i}$ problems as user math110 pointed out.

The condition $abcd = 1$ is quite annoying, we perform a technique called "homogenization". That is, the inequality presented is not homogeneous on both sides. We can make it homogeneous by multiplying a certain power of $P :=abcd = 1$ to each term: \begin{align*} aP^\frac{3}{4}+ b^2 P^\frac{2}{4} + cP^\frac{1}{4} + d^4 \ge a^{-1}P^\frac{5}{4} + b^{-2}P^\frac{6}{4} + c^{-3}P^\frac{7}{4} + d^{-4}P^\frac{8}{4} \end{align*} Note the pattern in choosing what power of $P$ to use. For each term of degree $k \in \{-4, -3, \cdots, 3, 4\}$, the power to raise $P$ is $(4 - k)/4$. To simplify some notation, denote $[k_1, k_2, k_3, k_4] = a^{k_1}b^{k_2}c^{k_3}d^{k_4}$. We can rewrite our inequality as \begin{align*} \left[\frac{7}{4}, \frac{3}{4}, \frac{3}{4}, \frac{3}{4}\right] + \left[\frac{1}{2}, \frac{5}{2}, \frac{1}{2}, \frac{1}{2}\right] + \left[\frac{1}{4}, \frac{1}{4}, \frac{13}{4}, \frac{1}{4}\right] + \left[0, 0, 0, 4\right] \\ \ge \left[\frac{1}{4}, \frac{5}{4}, \frac{5}{4}, \frac{5}{4}\right] +\left[\frac{3}{2}, -\frac{1}{2}, \frac{3}{2}, \frac{3}{2}\right] + \left[\frac{7}{4}, \frac{7}{4}, -\frac{5}{4}, \frac{7}{4}\right] + \left[\frac{8}{4}, \frac{8}{4}, \frac{8}{4}, -\frac{8}{4}\right] \end{align*} Transforming $(a, b, c, d) \mapsto (a^4, b^4, c^4, d^4)$ so we can work exclusively with integers while also preserving the order relation $0 \le a \le b \le c \le d$, \begin{align*} \left[7, 3, 3, 3\right] + \left[2, 10, 2, 2\right] + \left[1, 1, 13, 1\right] + \left[0, 0, 0, 16\right] \\ \ge \left[1, 5, 5, 5\right] +\left[6, -2, 6, 6\right] + \left[7, 7, -5, 7\right] + \left[8, 8, 8, -8\right] \end{align*} To further simplify notation, let us write $\ell_1 + \ell_2 + \ell_3 + \ell_4 \ge r_1 + r_2 + r_3 + r_4$, where each symbol represents the corresponding bracket notation. Now for some background. The technique of "isolated fudging" bounds each $r_j$ on one side of the inequality by a linear combination $\sum_{k=1}^{4}{w_{jk}\ell_k}$ such that positive weights $w_{jk}$ satisfy $\sum_{j=1}^{4}{w_{jk}} = 1$. So, we attempt to prove $\sum_{k=1}^{4}{w_{jk}\ell_k} \ge r_j$ and \begin{align*} \sum_{k=1}^{4}{\ell_k} =\sum_{k=1}^{4}\sum_{j=1}^{4}{w_{jk}\ell_k} = \sum_{j=1}^{4}\sum_{k=1}^{4}{w_{jk}\ell_k}\ge \sum_{j=1}^{4}{r_j} \end{align*} In our situation, we can explicitly solve for these coefficients through the series of equations \begin{align*} (r_j) = \sum_{k=1}^{4}{w_{jk}(\ell_k)} \end{align*} where $(\cdot)$ performed on each monomial denotes the vectorized versions with standard vector operations, e.g $(\ell_1) = (7, 3, 3, 3)$ as a vector. The reason we do this is because $r_j \le \sum_{k=1}^{4}{w_{jk}\ell_k}$ immediately follows from a weighted AM-GM with weights $w_{jk}$ and components $\ell_{k}$. If we are to solve all these equations, we get \begin{align*} (w_{jk}) = \frac{1}{25}\begin{pmatrix}-1 & 24 & 24 & 24 \\ 12 & -13 & 12 & 12 \\ 8 & 8 & - 17 & 8 \\ 6 & 6 & 6 & -19 \end{pmatrix} \end{align*} This is almost the answer, especially since the sum of each column equals 1. However, the problem is, weighted AM-GM only works when all the weights $w_{jk}$ are positive, so this proof seems a bit bogus. I argue performing the operation of moving values from one row of $(w_{jk})$ to another row, within the same column. This doesn't change the value of $\sum_{k=1}^{4}\sum_{j=1}^{4}{w_{jk}\ell_k}$, but it does change the lower-bound $r_j$. Somehow involve this with the order conditions $0 \le a \le b \le c \le d$ to arrive at the answer.

Answer 4

Suppose that it is true for $n=3$:

$$ \forall 0 \le a \le b \le c \wedge abc = 1 : a - \frac{1}{a} + b^2 - \frac{1}{b^2} + c^3 - \frac{1}{c^3} \ge 0. $$

Let $d \ge c \ge 1$, then it is clear that

$$ \forall 0 \le a \le b \le c \le d \wedge abcd = d : a - \frac{1}{a} + b^2 - \frac{1}{b^2} + c^3 - \frac{1}{c^3} + d^4 - \frac{1}{d^4} \ge 0. $$

Let us now define $$ a' = \frac{a}{\mu^{24}}, b' = \frac{b}{\mu^{12}}, c' = \frac{c}{\mu^{8}}, d' = \frac{d}{\mu^{6}}, $$ using $$ 24 = 4!, 12 = 4!/2, 8 = 4!/3, 6 = 4!/4. $$ We get $$ a' b' c' d' = \frac{a}{\mu^{24}} \frac{b}{\mu^{12}} \frac{c}{\mu^{8}} \frac{d}{\mu^{6}} = \frac{abcd}{\mu^{50}} = \frac{d}{\mu^{50}}, $$ so $$ a' b' c' d' =1, $$ if $$ \mu = \sqrt[50]{d}. $$ As $d \ge 1$, it is clear that $\mu \ge 1$. The inequality can be written as $$ \frac{a}{\mu^{24}} - \frac{\mu^{24}}{a} + \frac{b^2}{\mu^{24}} - \frac{\mu^{24}}{b^2} + \frac{c^3}{\mu^{24}} - \frac{\mu^{24}}{c^3} + \frac{d^4}{\mu^{24}} - \frac{\mu^{24}}{d^4} \ge 0. $$ Therefore $$ \mu^{24} \left( \frac{a}{\mu^{48}} - \frac{1}{a} + \frac{b^2}{\mu^{48}} - \frac{1}{b^2} + \frac{c^3}{\mu^{48}} - \frac{1}{c^3} + \frac{d^4}{\mu^{48}} - \frac{1}{d^4} \right) \ge 0. $$ Whence $$ \mu^{24} \left( a - \frac{1}{a} + b^2 - \frac{1}{b^2} + c^3 - \frac{1}{c^3} + d^4 - \frac{1}{d^4} \right) - \left( \mu^{12} - \frac{1}{\mu^{12}} \right)\big( a + b^2 + c^3 + d^4 \big) \ge 0. $$ Which can be written as $$ a - \frac{1}{a} + b^2 - \frac{1}{b^2} + c^3 - \frac{1}{c^3} + d^4 - \frac{1}{d^4} \ge \mu^{-24} \left( \mu^{12} - \frac{1}{\mu^{12}} \right)\big( a + b^2 + c^3 + d^4 \big). $$ As $\mu \ge 1$, we obtain

$$ \forall 0 \le a \le b \le c \le d \wedge abcd = 1 :\\ a - \frac{1}{a} + b^2 - \frac{1}{b^2} + c^3 - \frac{1}{c^3} + d^4 - \frac{1}{d^4} \ge 0. $$

This method allows to show that the case $n+1$ is true, IF the case $n$ is true.

Answer 5

Necessarily $\color{green}{d\ge 1}$ and at least one between $a,b,c$ is less or equal than $1$. We have the equivalent inequality $$a+b^2+c^3+d^4 - \frac{1}{a}-\frac{1}{b^2}-\frac{1}{c^3}-\frac{1}{d^4}\ge 0\qquad (*)$$ Arranging otherwise to compare the larger numbers involved with smaller ones $$(d^4-\frac 1a)+(a-\frac {1}{d^4}) +c^3+b^2-\frac{1}{c^3}-\frac{1}{b^2}\ge 0\qquad (**)$$ $$d^4-\frac 1a=d^4-bcd=d(d^3-bc)$$ Since $d^2\ge bc$ and $d\ge 1$ one has $d^3-bc$ is positive; it follows $$d^4-\frac 1a\ge 0\iff a-\frac{1}{d^4}\ge 0$$ therefore, if $b$ and $c$ are greater or equal than $1$ then the proof of $(**)$ is finished. Because of this we consider the restriction $\color{green}{0\lt a\le b\le c\lt 1\le d}$. $$*********************$$ Now, by the known $A-M\ge H-M$ and the statement’s inequality, one has the system $$\begin{cases}LHS\ge \frac {16}{RHS}\ge 0\\LHS\ge RHS\ge 0\end{cases} \Rightarrow (LHS)^2\ge 16\iff LHS\ge 4$$ Hence we have to show $$a+b^2+c^3+d^4\ge 4\qquad (***)$$ One has $$\frac{a^4+b^4+c^4+d^4}{4}\ge \sqrt[4]{a^4b^4c^4d^4}=1$$ But (since $\color{green}{0\lt a\le b\le c\lt 1}$) $$a\ge a^4\\b^2\ge b^4\\c^3\ge c^4\\d^4=d^4$$ Consequently $$a+b^2+c^3+d^4\ge a^4+b^4+c^4+d^4\ge 4$$ as we wanted to prove.