Prove $\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b} +\frac{81abc}{4(a+b+c)^2} \geqq \frac{7}{4} (a+b+c)$

Question

For $a,b,c>0$. Prove that$:$ $$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b} +\frac{81abc}{4(a+b+c)^2} \geqq \frac{7}{4} (a+b+c)$$

My proof:

We have$:$ $$\text{LHS}-\text{RHS} =\frac{g(a,b,c)}{4abc(a+b+c)^2} \geqq 0$$

Where
$g(a,b,c) =\frac{1}{16} \left( a+b \right) ^{2} \left( 2\,a+2\,b-c \right) ^{2} \left( a+b-2\,c \right) ^{2}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\frac{1}{64} \left( a-b \right) ^{2} \cdot \Big[ \left( 2\,c-a-b \right) ^{3} \left( 119\,a+119\,b+30\,c \right)$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\left( a+b-2\,c \right) ^{2} \left( 343\,{a}^{2}+346\,ab+343\,{b}^{2} \right) $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+24\, \left( 2\,c-a-b \right) \left( a+b \right) \left( 16\,{a}^{2}+a b+16\,{b}^{2} \right) $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+36\, \left( 4\,{a}^{2}-5\,ab+4\,{b}^{2} \right) \left( a+b \right) ^{ 2} \Big] \geqq 0$

which is clearly true for $c=\max\{a,b,c\}$

I wish to see another proof without $uvw$! Thanks for a real lot!

You can see also here.

Answer 1

Also, we can use the $uvw$'s technique.

Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, we need to prove that $$\frac{9v^4-6uw^3}{w^3}+\frac{81w^3}{36u^2}\geq\frac{21u}{4}$$ or $$w^6-5u^3w^3+4u^2v^4\geq0,$$ which is obviously true by $uvw$, which you don't want.

But we can use the following way.

$$(a-b)^2(a-c)^2(b-c)^2\geq0$$ it's $$3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6\geq0,$$ which gives $$3uv^2-2u^3-2\sqrt{(u^2-v^2)^3}\leq w^3\leq3uv^2-2u^3+2\sqrt{(u^2-v^2)^3}.$$ But for proving $w^6-5u^3w^3+4u^2v^4\geq0$ it's enough to prove that $$w^3\leq\frac{5u^3-\sqrt{25u^6-16u^2v^4}}{2},$$ for which it's enough to prove that $$3uv^2-2u^3+2\sqrt{(u^2-v^2)^3}\leq\frac{5u^3-\sqrt{25u^6-16u^2v^4}}{2}$$ or $$9u^3-6uv^2-4\sqrt{(u^2-v^2)^3}\geq\sqrt{25u^6-16u^2v^4}.$$ But by AM-GM $$9u^3-6uv^2=3u^3+6u(u^2-v^2)\geq6u^2\sqrt{u^2-v^2}\geq4\sqrt{(u^2-v^2)^3}.$$ Id est, it's enough to prove that: $$\left(9u^3-6uv^2-4\sqrt{(u^2-v^2)^3}\right)^2\geq25u^6-16u^2v^4$$ or $$18u^6-39u^4v^2+25u^2v^4-4v^6\geq6u(3u^2-2v^2)\sqrt{(u^2-v^2)^3}$$ or $$18u^4-21u^2v^2+4v^4\geq6u(3u^2-2v^2)\sqrt{u^2-v^2}$$ and since $$18u^4-21u^2v^2+4v^4=3(u^2-v^2)(6u^2-v^2)+v^4>0,$$ we need to prove that $$(18u^4-21u^2v^2+4v^4)^2\geq36u^2(3u^2-2v^2)^2(u^2-v^2)$$ or $$(3u^2-4v^2)^2\geq0.$$ Done!

Answer 2

here you go .

Answer 3

Here $a,b,c$ replaced for $x,y,z$ for convenience.

For $x,y,z>0$ prove that

\begin{align} \frac{xy}z+\frac{yz}x+\frac{zx}y+ \frac{81xyz}{4(x+y+z)^2} &\ge \tfrac74(x+y+z) \tag{1}\label{1} \end{align}

Using Ravi substitution,

\begin{align} x&=\rho-a ,\quad y=\rho-b ,\quad z=\rho-c , \end{align}

the inequality \eqref{1} transforms (omitting lengthy, but straightforward details) into equivalent inequality

\begin{align} 85\,r^2+32\,r\,R+64\,R^2 -15\,\rho^2 &\ge 0 \tag{2}\label{2} \end{align}

in terms of the semiperimeter $\rho$, inradius $r$ and circumradius $R$ of some valid triangle with the side lengths

\begin{align} a&=y+z ,\quad b=z+x ,\quad c=x+y . \end{align}

Dividing \eqref{2} by $R^2$, we get

\begin{align} 85\,v^2+32\,v+64 -15\,u^2 &\ge 0 \tag{3}\label{3} , \end{align} where $u=\rho/R$, $v=r/R$.

Now all we have to do is to check \eqref{3} for all valid triangles, that means for all $v\in[0,\tfrac12]$ and $u(v)\in[u_{\min}(v),u_{\max}(v)]$.

Obviously, \begin{align} 85\,v^2+32\,v+64 -15\,u^2 &\ge 85\,v^2+32\,v+64 -15\,u_{\max}(v)^2 \tag{4}\label{4} \end{align} where

\begin{align} u_{\max}&=\sqrt{27-(5-v)^2+2\sqrt{(1-2\,v)^3}} \tag{5}\label{5} , \end{align}

and we have \eqref{1} equivalent to

\begin{align} 50v^2-59v+17-15\sqrt{(1-2v)^3} &\ge 0 \tag{6}\label{6} ,\\ (50v^2-59v+17)^2-15^2((1-2v)^3) &\ge 0 \tag{7}\label{7} ,\\ (25v-8)^2(1-2v)^2 &\ge 0 \tag{8}\label{8} , \end{align} which holds for all values of $v\in[0,\tfrac12]$, that is for all valid triangles with the side lengths given by \eqref{2} and hence, \eqref{1} is true for all real $x,y,z$.

Answer 4

BW helps!

Let $a\leq b\leq c$, $b=a+u$ and $c=a+u+v.$

Thus, we need to prove that: $$9(u^2+uv+v^2)a^4+3(14u^3+21u^2v+3uv^2-2v^3)a^2+$$ $$+(73u^4+146u^3v+66u^2v^2-7uv^3+v^4)a^2+$$ $$+u(56u^4+140u^3v+114u^2v^2+31uv^3+v^4)a+$$ $$+4u^2(2u^2+3uv+v^2)^2\geq0,$$ for which it's enough to prove that: $$4(u^2+uv+v^2)(73u^4+146u^3v+66u^2v^2-7uv^3+v^4)\geq(14u^3+21u^2v+3uv^2-2v^3)^2$$ or $$292u^6+876u^5v-85u^4v^2+610u^3v^3+391u^2v^4+8uv^5+20v^6\geq0,$$ which is obvious.

Answer 5

I found a proof by Titu's Lemma$:$

Let $$\text{LHS} -\text{RHS} \equiv \frac{g(a,b,c)}{4abc(a+b+c)^2}$$

But we have

\begin{align*} g(a,\,b,\,c) &=\frac{1}{2} \sum\limits_{cyc} c^2(a+b-5c)^2 (a-b)^2-9(a-b)^2(b-c)^2(c-a)^2 \\&\geq \frac{1}{2(a^2+b^2+c^2)} \Big[\sum\limits_{cyc} c^2(a+b-5c)(a-b)\Big]^2-9(a-b)^2(b-c)^2(c-a)^2\\&={\frac { \left( a-b \right) ^{2} \left( b-c \right) ^{2} \left( c-a \right) ^{2} \left( 7\,{a}^{2}+50\,ab+50\,ac+7\,{b}^{2}+50\,bc+7\,{c} ^{2} \right) }{2\,(a^2+b^2+c^2)}} \geq 0\end{align*}

So we are done.