# $\sqrt{x^2+x}+\sqrt{y^2+y}+\sqrt{z^2+z}+\sqrt{w^2+w}\le(x+1)(y+1)(z+1)(w+1)$

When $x,y,z,w$ is positive, prove that$\sqrt{x^2+x}+\sqrt{y^2+y}+\sqrt{z^2+z}+\sqrt{w^2+w}\le(x+1)(y+1)(z+1)(w+1)$

This inequality is a repost of now deleted MSE question.

What I tried: By C-S,$$(LHS)^2=\left(\sum_{cyc}{\sqrt{x^2+x}}\right)^2\le\sum_{cyc}x\sum_{cyc}(x+1)=\sum_{cyc}x\left(4+\sum_{cyc}x\right)$$ and it is left to prove $(x+y+z+w)(x+y+z+w+4)\le((x+1)(y+1)(z+1)(w+1))^2$.

It is only true for $x+y+z+w\le\frac{1}{2}$.

On the other hand, $$LHS=\sum_{cyc}{\sqrt{x^2+x}}<\sum_{cyc}{\sqrt{x^2+x+0.25}}\le(x+1)(y+1)(z+1)(w+1)$$if $1\le xyzw+xyz+xyz+xzw+yzw+xy+yz+zw+wx+xz+yw$.

However, I feel like this kind of approach will always leave hole and will not be sufficient for the problem. Any solutions?

• From where does this problem come? Jul 8, 2018 at 13:18
• I don't know, someone previously posted this inequality and was deleted for not showing any effort or context. Jul 8, 2018 at 13:21
• Can you send me a link please? Jul 8, 2018 at 13:35
• This is the link of deleted question (I cannot see its content now): math.stackexchange.com/questions/2795656/… Jul 8, 2018 at 13:42
• i arrived at the same inequality like you i used AM-GM Jul 8, 2018 at 13:44

Lemma 1. The inequality holds when one of variables is $$0$$.

We will prove this lemma later, with the similar argument of the inequality itself.

Lemma 2. The inequality holds when one of variables is at least $$3$$.

Without loss of generality, let $$x\ge3$$. It is easy to see $$\left(3y+\frac16\right)^2>y^2+y$$, therefore$$\sqrt{x^2+x}+\sqrt{y^2+y}+\sqrt{z^2+z}+\sqrt{w^2+w}< x+\frac12+3y+\frac16+3z+\frac16+3w+\frac16\\\le x+xy+xz+xw+1\le\left(1+x\right)\left(1+y\right)\left(1+z\right)\left(1+w\right)$$

Now, let $$f\left(x,y,z,w\right)=\left(1+x\right)\left(1+y\right)\left(1+z\right)\left(1+w\right)-\sum_{cyc}\sqrt{x^2+x}$$. When we want to find global minimum of multivatiable function, we need to find critical points and boundary points. By lemma 1 and 2, it is enough to show for critical points. Now, notice that$$\frac{d}{dx}f\left(x,y,z,w\right)=0\iff\left(1+y\right)\left(1+z\right)\left(1+w\right)=\frac{2x+1}{2\sqrt{x^2+x}}\\\iff\left(1+x\right)\left(1+y\right)\left(1+z\right)\left(1+w\right)=\frac{\left(x+1\right)\left(2x+1\right)}{2\sqrt{x^2+x}}$$ Let $$g\left(x\right)=\frac{\left(x+1\right)\left(2x+1\right)}{2\sqrt{x^2+x}}$$. At critical points of $$f$$, $$g\left(x\right)=g\left(y\right)=g\left(z\right)=g\left(w\right).$$ Since $$g''\left(x\right)=\frac{\left(x + 1\right) \left(2 x + 3\right)}{8 \left(x^2+x\right)^{5/2}}>0$$, there are at most two solutions for $$g\left(x\right)=c$$ for any positive $$c$$. Therefore, we may assume $$x=z=w$$ or $$x=z\land y=w$$.

Case 1. $$x=z=w$$, prove $$3\sqrt{x^2+x}+\sqrt{y^2+y}\le\left(x+1\right)^3\left(y+1\right)$$

If $$y\ge\frac12$$, then $$3\sqrt{x^2+x}+\sqrt{y^2+y}\le3\sqrt{x^2+x}+y+\frac12$$, therefore we need to show $$\left(x+1\right)^3\left(y+1\right)-3\sqrt{x^2+x}-y-\frac12\ge0$$. Since it is increasing linear function of $$y$$, it is enough to show for $$y=\frac12$$, in other words, $$\frac32\left(x+1\right)^3-3\sqrt{x^2+x}-1\ge0$$. Notice that from $$9 x^6 + 54 x^5 + 135 x^4 + 168 x^3 + 63 x^2 - 18 x + 1\ge0$$ for all positive $$x$$, $$\left(3\left(x+1\right)^3-2\right)^2\ge36\left(x^2+x\right)$$, therefore our claim is proved.

If $$y\le\frac12$$, then notice that $$\left(x^2+x+\frac14\right)^2-\left(x^2+x\right)=\left(x^2+x-\frac14\right)^2\ge0$$. We need to show $$\left(x+1\right)^3\left(y+1\right)-3\sqrt{x^2+x}-y^2-y-\frac14\ge0$$. Since it is concave function of $$y$$, it is enough to prove for extreme values of $$y$$, which is $$0$$ and $$\frac12$$. For $$y=\frac12$$, it is $$\frac32\left(x+1\right)^3-3\sqrt{x^2+x}-1\ge0$$ which we already proved. For $$y=0$$, it is $$\left(x+1\right)^3-3\sqrt{x^2+x}-\frac14\ge0$$. Again from $$x^2+x+\frac14\ge\sqrt{x^2+x}$$, $$\left(x+1\right)^3-3\sqrt{x^2+x}-\frac14\ge\left(x+1\right)^3-3\left(x^2+x+\frac14\right)-\frac14=x^3\ge0$$

Case 2. $$x=z$$ and $$y=w$$, prove $$2\sqrt{x^2+x}+2\sqrt{y^2+y}\le\left(x+1\right)^2\left(y+1\right)^2$$ Notice that$$\left(2x+1\right)^2\left(2y+1\right)^2=\left(2\sqrt{x^2+x}+2\sqrt{y^2+y}\right)^2+\left(4\sqrt{x^2+x}\sqrt{y^2+y}-1\right)^2$$Therefore $$2\sqrt{x^2+x}+2\sqrt{y^2+y}\le\left(2x+1\right)\left(2y+1\right)<\left(x+1\right)^2\left(y+1\right)^2$$.

Now we checked all critical points of $$f$$, so we finish the proof.

Proof of lemma 1. We want to find global minimum of $$f_1\left(x,y,z\right)=f\left(x,y,z,0\right)$$. We already checked boundary points where $$x,y,z$$ is sufficiently large. For one of variable is $$0$$, WLOG assume $$z=0$$. $$\sqrt{x^2+x}+\sqrt{y^2+y}\le x+\frac12+y+\frac12\le\left(x+1\right)\left(y+1\right)$$Therefore, we only need to check critical points of $$f_1$$. By the same argument as main proof, at critical points of $$f_1$$, $$g\left(x\right)=g\left(y\right)=g\left(z\right)$$ and at least two variables are same. We can assume $$x=z$$.

Now we need to show $$2\sqrt{x^2+x}+\sqrt{y^2+y}\le\left(x+1\right)^2\left(y+1\right)$$. Indeed,$$\left(2x+1\right)^2\left(y+1\right)^2=\left(2\sqrt{x^2+x}+\sqrt{y^2+y}\right)^2+\left(2\sqrt{x^2+x}\sqrt{y^2+y}-1\right)^2+4xy\left(x+1\right)+y+1$$ therefore $$2\sqrt{x^2+x}+\sqrt{y^2+y}\le\left(2x+1\right)\left(y+1\right)\le\left(x+1\right)^2\left(y+1\right)$$.