Prove that $\frac{m}{\sqrt{7m+2}} + \frac{n}{\sqrt{7n+2}} + \frac{p}{\sqrt{7p+2}} \ge 1.$

Question

Let $m,n,p \ge 0$ such that $m+n+p = mn+np+pm > 0$. Prove that:

$$\frac{m}{\sqrt{7m+2}} + \frac{n}{\sqrt{7n+2}} + \frac{p}{\sqrt{7p+2}} \ge 1.$$

The equality occurs when $(m,n,p)=(1,1,1)$ or $(m,n,p)=(2,2,0)$

I tried to use Holder inequality:

$$\left(\sum\frac{m}{\sqrt{7m+2}}\right)^2\left(\sum m(7m+2)(4m+n+p)^3\right) \ge \left(\sum m(4m+n+p)\right)^3$$ Thus, I have to prove: $$\left(\sum m(4m+n+p)\right)^3 \ge \left(\sum m(7m+2)(4m+n+p)^3\right) $$

But this seems to be a buffalo way, can anyone help me with this step. Thank you a lots.

Answer 1

Problem. Let $x, y, z\ge 0$ with $x + y + z = xy + yz + zx > 0$. Prove that $$\frac{x}{\sqrt{7x+2}} + \frac{y}{\sqrt{7y+2}} + \frac{z}{\sqrt{7z+2}} \ge 1.$$

Alternative proof.

By AM-GM, we have $\frac{x}{\sqrt{7x+2}} \ge \frac{2x}{\frac{7x+2}{x+2} + (x+2)}$ etc. It suffices to prove that $$\sum_{\mathrm{cyc}} \frac{2x}{\frac{7x+2}{x+2} + (x+2)} \ge 1. \tag{1}$$

Let $p := x + y + z, q := xy + yz + zx, r := xyz$. The condition $x + y + z = xy + yz + zx$ becomes $p = q$. Using $p = q \le p^2/3$, we have $p \ge 3$.

Using $q = p$, (1) is equivalent to $$5r^2 + (210p - 149)r + 18(8p + 3)(p - 4) \ge 0. \tag{2}$$

If $p \ge 4$, clearly (2) is true.

If $3 \le p < 4$, using degree three Schur inequality $r \ge \frac{4pq - p^3}{9} = \frac{4p^2 - p^3}{9}\ge 0$, it suffices to prove that $$5\left( \frac{4p^2 - p^3}{9}\right)^2 + (210p - 149)\cdot \frac{4p^2 - p^3}{9} + 18(8p + 3)(p - 4) \ge 0,$$ or $$\frac{1}{81}(p - 3)(4 - p)(-5p^4 + 5p^3 + 1905p^2 + 4374p + 1458) \ge 0$$ which is true.

We are done.

Answer 2

Another way.

Firstly we'll prove the following stronger inequality.

Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc+abc=4$. Prove that: $$\frac{a}{\sqrt{7a+2}}+\frac{b}{\sqrt{7b+2}}+\frac{c}{\sqrt{7c+2}}\geq1.$$

Indeed, let $a=\frac{2x}{y+z}$ and $b=\frac{2y}{x+z},$ where $x$, $y$ and $z$ are non-negatives such that $xy+xz+yz\neq0.$

Thus, from the condition $c=\frac{2z}{x+y}$ and we need to prove that: $$\sum_{cyc}\frac{\frac{2x}{y+z}}{\sqrt{\frac{14x}{y+z}+2}}\geq1$$ or $$\sum_{cyc}\frac{x}{\sqrt{(y+z)(7x+y+z)}}\geq\frac{1}{\sqrt2}.$$ Now, by Holder $$\left(\sum_{cyc}\frac{x}{\sqrt{(y+z)(7x+y+z)}}\right)^2\sum_{cyc}x(y+z)(7x+y+z)\geq(x+y+z)^3$$ and it's enough to prove that: $$2(x+y+z)^3\geq\sum_{cyc}x(y+z)(7x+y+z)$$ or $$\sum_{cyc}(x^3-x^2y-x^2z+xyz)\geq0,$$ which is true by Schur.

Now, easy to see that $a+b+c\geq ab+ac+bc$ because it's $$\sum_{cyc}\frac{x}{y+z}\geq\sum_{cyc}\frac{2yz}{(x+z)(x+y)},$$ which is $$\sum_{cyc}x(x+y)(x+z)\geq2\sum_{cyc}yz(y+z)$$ or $$\sum_{cyc}(x^3-x^2y-x^2z+xyz)\geq0,$$ which is Schur again.

Id est, $1\geq\frac{ab+ac+bc}{a+b+c}$ and $$\sum_{cyc}\frac{a}{\sqrt{7a\cdot\frac{ab+ac+bc}{a+b+c}+2\left(\frac{ab+ac+bc}{a+b+c}\right)^2}}\geq\sum_{cyc}\frac{a}{\sqrt{7a+2}}\geq1,$$ which gives $$\sum_{cyc}\frac{a}{\sqrt{7a\cdot\frac{ab+ac+bc}{a+b+c}+2\left(\frac{ab+ac+bc}{a+b+c}\right)^2}}\geq1,$$ which is homogeneous already.

Thus, we can assume $a+b+c=ab+ac+bc$ and we obtain your inequality.

Answer 3

Another way.

Let $m+n+p=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, the condition does not depend on $w^3$ and gives $u=v^2\geq1$ Now, by AM-GM $$\sum_{cyc}\frac{m}{\sqrt{7m+2}}=\sqrt{\sum_{cyc}\left(\frac{m^2}{7m+2}+\frac{2mn}{\sqrt{(7m+2)(7n+2)}}\right)}\geq\sqrt{\sum_{cyc}\left(\frac{m^2}{7m+2}+\frac{4mn}{7m+7n+4}\right)}$$ and it's enough to prove that: $$\sum_{cyc}\left(\frac{m^2}{7m+2}+\frac{4mn}{7m+7n+4}\right)\geq1$$ or $$\tfrac{147uw^3-42w^3+126uv^2+36u^2-24v^2}{343w^3+294v^2+84u+8}+\tfrac{4(147uw^3+84w^3+441u^2+252uv^2+48v^2)}{-343w^3+1764u^2+3087uv^2+672u+558v^2+64}\geq1$$ or $f(w^3)\geq0,$ where $$f(w^3)=(147uw^3-42w^3+126uv^2+36u^2-24v^2)(-343w^3+1764u^2+3087uv^2+672u+558v^2+64)+$$ $$4(147uw^3+84w^3+441u^2+252uv^2+48v^2)(343w^3+294v^2+84u+8)-$$ $$(-343w^3+1764u^2+3087uv^2+672u+558v^2+64)(343w^3+294v^2+84u+8).$$ But by Schur $$f'(w^3)=(147w^3-42)(-343w^3+1764u^2+3087uv^2+672u+558v^2+64)+$$ $$+(147uw^3-42w^3+126uv^2+36u^2-24v^2)(-343)+$$ $$4(147u+84)(343w^3+294v^2+84u+8)+4(147uw^3+84w^3+441u^2+252uv^2+48v^2)343-$$ $$(-343)(343w^3+294v^2+84u+8)-(-343w^3+1764u^2+3087uv^2+672u+558v^2+64)343=$$ $$=259308u^3+453789u^2v^2+61740u^2-624661uv^2+302526uw^3-187572u+47334v^2+494606w^3-19208\geq$$ $$\geq259308u^3+453789u^3+61740u^2-624661u^2-187572u+47334u+9270w^3-19208\geq$$ $$\geq259308u^3+453789u^3+61740u^2-624661u^2-187572u+47334u+9270(4uv^2-3u^3)-19208=$$ $$=(u-1)(685287u^2+159446u+19208)\geq0,$$ which says that $f$ increases.

Thus, by $uvw$ it's enough to prove $$\sum_{cyc}\left(\frac{m^2}{7m+2}+\frac{4mn}{7m+7n+4}\right)\geq1$$ in the following cases.

1. $w^3=0$.

Let $p=0$ and $m+n=x$.

Thus, $x=m+n=mn\leq\left(\frac{m+n}{2}\right)^2=\frac{x^2}{4},$ which gives $x\geq4$ and we need to prove here that $$\frac{m^2}{7m+2}+\frac{n^2}{7n+2}+\frac{4mn}{7m+7n+4}\geq1$$ or $$\frac{7x^2+2x^2-4x}{49x+14x+4}+\frac{4x}{7x+4}\geq1$$ or $$(x-4)(63x^2+71x+4)\geq0.$$ 2. Two variables are equal.

Let $m=n$.

Thus, from the condition $p=\frac{m(2-m)}{2m-1}$, where $\frac{1}{2}<m\leq2$ and we need to prove that: $$\frac{2m^2}{7m+2}+\frac{\left(\frac{m(2-m)}{2m-1}\right)^2}{\frac{7m(2-m)}{2m-1}+2}+\frac{4m^2}{14m+4}+\frac{8m\cdot\frac{m(2-m)}{2m-1}}{7m+\frac{7m(2-m)}{2m-1}+4}\geq1$$ or $$(2-m)(m-1)^2(441m^4-1015m^3+116m^2+70m-8)\leq0,$$ which is true for $\frac{1}{2}<m\leq2$.

About $uvw$ see here: https://artofproblemsolving.com/community/c6h278791

Answer 4

Rewriting @Michael Rozenberg's nice solution:

Proof. We claim that $mn + np + pm + mnp \ge 4$. Indeed, letting $P := m + n + p, Q := mn + np + pm, R := mnp$, we have $P = Q > 0$. Using $P^2 \ge 3Q$, we have $P \ge 3$. If $P > 4$, we have $Q + R \ge Q \ge 4$. If $3 \le P \le 4$, using degree three Schur inequality $R \ge \frac{4PQ - P^3}{9}$, we have $Q + R - 4 \ge Q + \frac{4PQ - P^3}{9} - 4 = \frac19(P + 3)(P-3)(4-P)\ge 0$. The claim is proved.

To proceed, we need the following auxiliary result.

Fact 1. Let $a, b, c\ge 0$ with $ab + bc + ca + abc \ge 4$. Then $\frac{a}{\sqrt{7a+2}}+\frac{b}{\sqrt{7b+2}}+\frac{c}{\sqrt{7c+2}}\geq1$.
(The proof of Fact 1 is given at the end.)

Now, by Fact 1, letting $a = m, b = n, c = p$, the desired result follows. We are done.

$\phantom{2}$

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Proof of Fact 1.

Let $f(k) := \frac{ab + bc + ca}{k^2} + \frac{abc}{k^3} - 4$. We have $f(1) \ge 0$ and $f(\infty) = -4$. Thus, there exists $k \ge 1$ such that $\frac{ab + bc + ca}{k^2} + \frac{abc}{k^3} - 4 = 0$. Let $u := \frac{a}{k}, v := \frac{b}{k}, w := \frac{c}{k}$. We have $uv + vw + wu + uvw = 4$.

We use the known parameterization for $uv + vw + wu + uvw = 4$: $$u = \frac{2x}{y + z}, \quad v = \frac{2y}{z + x}, \quad w = \frac{2z}{x + y}$$ for $x, y, z \ge 0$ with $(x + y)(y+z)(z+x) > 0$.

We have \begin{align*} &\frac{a}{\sqrt{7a+2}}+\frac{b}{\sqrt{7b+2}}+\frac{c}{\sqrt{7c+2}}\\ ={}& \frac{ku}{\sqrt{7ku+2}}+\frac{kv}{\sqrt{7kv+2}}+\frac{kw}{\sqrt{7kw+2}}\\ ={}& \frac{u}{\sqrt{7u/k + 2/k^2}}+\frac{v}{\sqrt{7v/k + 2/k^2}}+\frac{w}{\sqrt{7w/k + 2/k^2}}\\ \ge{}& \frac{u}{\sqrt{7u + 2}}+\frac{v}{\sqrt{7v + 2}}+\frac{w}{\sqrt{7w + 2}}\\ ={}& \frac{x\sqrt{2}}{\sqrt{(y + z)(7x + y + z)}} + \frac{y\sqrt{2}}{\sqrt{(z + x)(7y + z + x)}} + \frac{z\sqrt{2}}{\sqrt{(x + y)(7z + x + y)}}. \end{align*}

It suffices to prove that $$\sum_{\mathrm{cyc}} \frac{x}{\sqrt{(y + z)(7x + y + z)}} \ge \frac{1}{\sqrt{2}}$$

Then we use Holder inequality. See @Michael Rozenberg's answer for more details.

Answer 5

Here's the correct Holder's bacteria that saves the equality occurring: