Asymmetric inequality in three variables $\frac{3(a+b)^2(b+c)^2}{4ab^2c} \geq 7+\frac{5(a^2+2b^2+c^2)}{(a+b)(b+c)}$

Question

Consider three real positive variables $a,\ b$ and $c$. Prove that the following inequality holds:

$$\frac{3(a+b)^2(b+c)^2}{4ab^2c} \geq 7+\frac{5(a^2+2b^2+c^2)}{(a+b)(b+c)}$$

My progress: We can prove that both sides are greater than $12$ using AM-GM:

$$LHS \geq \frac{3 \cdot 4ab \cdot 4bc}{4a^2bc} = 12$$

and

$$RHS \geq 7+\frac{5[(a+b)^2+(b+c)^2]}{2(a+b)(b+c)} \geq 7+5 = 12$$

So, substract $12$ from both sides and write the inequality into:

$$3\cdot \frac{(a+b)^2(b+c)^2-16ab^2c}{4ab^2c} \geq 5 \cdot \frac{a^2+b^2+c^2-ab-bc-ca}{(a+b)(b+c)}$$

or

$$3\cdot \frac{(b+c)^2(a-b)^2+4ab(b-c)^2}{4ab^2c} \geq \frac{5}{2}\cdot \frac{(a-b)^2+(b-c)^2+(c-a)^2}{(a+b)(b+c)}$$

My next idea was to use $(c-a)^2\leq 2[(a-b)^2+(b-c)^2]$ and write it into a sum of square form with only $(a-b)^2$ and $(b-c)^2$. However, I couldn't reach significant progress.

Answer 1

Partial solution. I hope, it can help.

Let $a+c=2p$ and $ac=q^2,$ where $q>0.$

Thus, by AM-GM $p\geq q$ and we need to prove that $$\frac{3(b^2+2bp+q^2)^2}{4q^2b^2}\geq7+\frac{5(b^2+4p^2-2q^2)}{b^2+2pb+q^2}.$$ Now, consider two cases:

1. $b\geq q$, $p=q+u$ and $b=q+v$.

Thus, we need to prove that: $$72uq^5+4(16u^2+94uv+19v^2)q^4+8(3u^3+34u^2v+85uv^2+19v^3)q^3+$$ $$+4v(18u^3+97u^2v+130uv^2+28v^3)q^2+18v^2(2u+v)^2(u+2v)q+3v^3(2u+v)^3\geq0,$$ which is obviously true.

2. $b\leq q$, $q=b+u$, $p=b+u+v$.

Thus, we need to prove that: $$72vb^5+4(19u^2-4uv+16v^2)b^4+8(19u^3-13u^2v-2uv^2+3v^3)b^3+$$ $$+4u^2(28u^2-4uv-11v^2)b^2+18u^4(2u+v)b+3u^6\geq0.$$ Now, we see that it's a cubic inequality of $v$ and after using a derivative we can get a minimal point and to end a proof.

Answer 2

Here is a (somewhat) complete, but cumbersome, method:

Let $$f(a,b,c)=3 a^3 b^3-11 a^3 b^2 c+9 a^3 b c^2+3 a^3 c^3+9 a^2 b^4-a^2 b^3 c-a^2 b^2 c^2+9 a^2 b c^3+9 a b^5-41 a b^4 c-a b^3 c^2-11 a b^2 c^3+3 b^6+9 b^5 c+9 b^4 c^2+3 b^3 c^3.$$ Then the inequality is the same as $$\frac{f(a,b,c)}{4 a b^2 (a + b) c (b + c)}\geq0.$$ By positivity of $a,b,c$, this is equivalent to $f(a,b,c)\geq0$. For $\lambda\in\mathbb R$ it is easy to check that $$f(\lambda a,\lambda b,\lambda c)=\lambda^6 f(a,b,c).$$ So without loss of generality it is enough to prove that $$g(a,b)\overset{\text{Def.}}=f(a,b,1)=a^3 (b (b (3 b-11)+9)+3)+a^2 b (b+1) (b (9 b-10)+9)+a b^2 (b (b (9 b-41)-1)-11)+3 b^3 (b+1)^3\geq0$$ for all $a,b\geq0$.

Proving this turns out to be very hard. So here is the best I could come up with: It is clear that $g(a,b)\geq0$ once $a,b$ are both large enough (since the terms $a b^5$ and $a^2 b^4$ will dominate all the negative terms) or if $a=0$ or if $b=0$. So we only need to check the points at which the gradient of $g$ vanishes. Using Mathematica (I wasn't able to do this manually), we find that for $a,b\geq0$ we have $$\nabla g(a,b)\iff (b=1\land a=1)\lor \left(P(b)=0\land a=-\frac{29669214164675939369671 b^{12}}{37219329419297521945472}-\frac{2690529003579005413929 b^{11}}{58705566907409340608}-\frac{224178267577049216215212647 b^{10}}{837434911934194243773120}-\frac{190470719842608496758497689 b^9}{4187174559670971218865600}+\frac{7842414818041358706380789953 b^8}{25123047358025827313193600}-\frac{2318470162882413536669733803 b^7}{2093587279835485609432800}+\frac{724942721587695792585411601 b^6}{232620808870609512159200}-\frac{5558706902918498564333817371 b^5}{2093587279835485609432800}+\frac{18299666529001271785024272871 b^4}{8374349119341942437731200}-\frac{8020692061649137411375535189 b^3}{4187174559670971218865600}+\frac{9349011842755660085287755043 b^2}{12561523679012913656596800}-\frac{1884408330289162109628656449 b}{4187174559670971218865600}+\frac{4871420917897848040148437}{37219329419297521945472}\right),$$

where $P(b)=-9 + 14 b + 8 b^2 + 54 b^3 + 9 b^4$ has exactly one positive real root. Indeed for the first solution we have $g(a,b)=0$ and for the second we have by numerical methods $$g(a,b)\approx 0.426984\geq0$$ which implies that $g\geq0$ as required.

Answer 3

Making the coordinates transformation

$$ \cases{ \frac{a+b}{a}=x\\ \frac{b+c}{c}=y\\ a b^2 c=z } $$

solving for $a,b,c$ we have

$$ \left\{ \begin{array}{rcl} a & = & \frac{z}{(y-1)^2 \left(\frac{(x-1) z}{(y-1)^3}\right)^{3/4}} \\ b & = & (y-1) \sqrt[4]{\frac{(x-1) z}{(y-1)^3}} \\ c & = & \sqrt[4]{\frac{(x-1) z}{(y-1)^3}} \\ \end{array} \right. $$

conditioned to $x > 1,\ y > 1,\ z>0$. Substituting into

$$ 3\frac{(a+b)^2(b+c)^2}{4ab^2c}- 7 - 5\frac{a^2+2b^2+c^2}{(a+b)(b+c)}\ge 0 $$

we have

$$ f(x,y) = \frac{3 x^3 y^3-4 x^2 (y (17 y-27)+15)+4 x (y (27 y-47)+30)-20 (3 (y-2) y+4)}{4 (x-1) x (y-1) y}\ge 0 $$

Now $f(x,y)$ has a minimum for $x=y=2$ such that $f(2,2) = 0$

Follows a plot of $f(x,y)$

Answer 4

Proof: Due to homogeneity, assume that $c = 1$. After clearing the denominators, it suffices to prove that \begin{align} &(3 b^3-11 b^2+9 b+3) a^3+(9 b^4-b^3-b^2+9 b) a^2\\ &\qquad +(9 b^5-41 b^4-b^3-11 b^2) a+3 b^6+9 b^5+9 b^4+3 b^3 \ge 0. \end{align} For each fixed $b > 0$, we need to prove that $f(a)\ge 0$ for all $a > 0$, where \begin{align} f(a) &= (3 b^3-11 b^2+9 b+3) a^3+(9 b^4-b^3-b^2+9 b) a^2\\ &\qquad +(9 b^5-41 b^4-b^3-11 b^2) a+3 b^6+9 b^5+9 b^4+3 b^3. \end{align} We split into two cases:

1) $b = 1$: We have $f(a) = 4(a+6)(a-1)^2\ge 0$. True.

2) $0 < b < 1$ or $b > 1$: Since $3 b^3-11 b^2+9 b+3 = 3(b-2)^2b + (b-\frac{3}{2})^2 + \frac{3}{4} > 0$, $f$ is cubic. The discriminant of $f$ is $\mathrm{discr}(f) = -48(b-1)^2 b^6g(b)$ where \begin{align} g(b) &= 18225 b^8-34992 b^7+15772 b^6-1552 b^5+26942 b^4\\ &\quad +11888 b^3+6380 b^2+2000 b+25. \end{align} We can prove that $g(b) > 0$ (the proof is give later). Thus, $\mathrm{discr}(f) < 0$. Thus, $f(a) = 0$ has exactly one real root on $(-\infty, +\infty)$. Note also that $f(-\infty) = -\infty$, $f(+\infty) = + \infty$, and $f(0) = 3 b^6+9 b^5+9 b^4+3 b^3 > 0$. Thus, $f(a) > 0$ for all $a > 0$. We are done.

$\phantom{2}$

Proof of $g(b)>0$: If $0 < b < 1$, then \begin{align} g(b) &= 18225b^8+15772b^6+2286b^4+6380b^2+2000b+25\\ &\qquad + 16b^3(1-b)(2187b^3+2187b^2+2284b+743)\\ &> 0. \end{align} If $b > 1$, we have \begin{align} g(b) &= 18225 (b-1)^8+110808 (b-1)^7+281128 (b-1)^6\\ &\qquad +378848 (b-1)^5 +306792 (b-1)^4 +215456 (b-1)^3\\ &\qquad +200224 (b-1)^2+145920(b-1) + 44688\\ &> 0. \end{align} We are done.

Answer 5

Another proof:

Since the inequality is symmetric in $a$ and $c$, WLOG, assume that $a \ge c$. Due to homogeneity, assume that $c = 1$. Let $a = 1 + s$ for $s \ge 0$.

We split into two cases:

1) $0 < b \le 1$: Let $b = \frac{1}{1+t}$ for $t \ge 0$. We have \begin{align} \mathrm{LHS} - \mathrm{RHS} &= \frac{1}{4(2+t)(st+s+t+2)(1+s)(1+t)^2}f(s,t) \end{align} where \begin{align} f(s,t) &= 3 s^3 t^6+27 s^3 t^5+9 s^2 t^6+79 s^3 t^4+90 s^2 t^5+9 s t^6+109 s^3 t^3\\ &\quad +281 s^2 t^4+99 s t^5+3 t^6+78 s^3 t^2+412 s^2 t^3+314 s t^4+36 t^5\\ &\quad +28 s^3 t+324 s^2 t^2+452 s t^3+112 t^4+4 s^3+140 s^2 t+304 s t^2\\ &\quad +152 t^3+28 s^2+76 s t+76 t^2. \end{align} Clearly, $f(s,t) \ge 0$. True.

2) $b > 1$: Let $b = 1+r$ for $r > 0$. We have \begin{align} \mathrm{LHS} - \mathrm{RHS} &= \frac{1}{4(1+s)(1+r)^2(2+s+r)(2+r)}g(s,r) \end{align} where \begin{align} g(s, r) &= 3 r^6+9 r^5 s+9 r^4 s^2+36 r^5+22 r^4 s+44 r^3 s^2\\ &\quad +112 r^4+4 r^3 s+44 r^2 s^2+76 r^3+9 r s^2+9 s^2+\tfrac{15}{16} s^3\\ &\quad + 19 (2 r-s)^2+19 r (2 r-s)^2+3 (r-1)^2 r s^3+4 s^3 (r-\tfrac78)^2. \end{align} Clearly, $g(s,r) > 0$. True.

We are done.