Prove that for any $x \ge 0, y\ge 0$ the inequality $\frac12(x+y)^2+\frac14(x+y) \ge x \sqrt y + y\sqrt x$ holds

Question

Prove that for any $x \ge 0$, $y\ge 0$, the inequality $$\frac12(x+y)^2+\frac14(x+y) \ge x \sqrt y + y\sqrt x$$ is true.

Okay, so I thought of the inequality of the means. Not just about the arithmetic and the geometric, but about the harmonic and the quadratic mean. But where can I use them and how, I don't really know. I also thought of showing that

$$\frac12(x+y)^2 + \frac14(x+y) - (x \sqrt y + y \sqrt x) > 0$$

Maybe by assumption and contradicting the assumption. All the ideas are welcome! Thank you!!

Answer 1

I have an idea! We want to show $$2(x+y)^2 + (x+y) \ge 4(x\sqrt y + y\sqrt x)$$ i.e., $$2x^2 + 2y^2 + 4xy + x + y \ge 4(x\sqrt y + y\sqrt x)$$ First of all, using the AM-GM inequality, $$2x^2 + 2y^2 \ge 2\cdot \sqrt{4x^2y^2} = 4xy$$ so that $$2x^2 + 2y^2 + y \ge 4xy + y \tag{1}$$ Using the AM-GM inequality again, $$4xy + y \ge 2\sqrt{4xy^2} = 4y\sqrt x \tag{2}$$ Also, $$4xy + x \ge 4x\sqrt y \tag{3}$$ From $(1)$ and $(2)$, we have $$2x^2 + 2y^2 + y \ge 4y\sqrt{x} \tag{4}$$ The desired inequality follows by adding $(3)$ and $(4)$.

Answer 2

It sufficies to prove that :

$$ \begin{align}2xy+\frac 14(x+y)&≥\sqrt {xy}(\sqrt x+\sqrt y)\end{align} $$

by AM-GM inequality.

Then, it sufficies to prove that :

$$ \begin{align}2xy+\frac 18\left(\sqrt x+\sqrt y\right)^2&≥\sqrt {xy}\left(\sqrt x+\sqrt y\right)\end{align} $$

by Cauchy-Schwars inequality.

which is correct, since

$$ \begin{align}16xy+\left(\sqrt x+\sqrt y\right)^2-8\sqrt {xy}\left(\sqrt x+\sqrt y\right)=\\ \left(4\sqrt {xy}-\left(\sqrt x+\sqrt y\right)\right)^2≥0 \thinspace.\end{align} $$

Answer 3

We have that

$$\frac12(x+y)^2+\frac14(x+y) \ge x \sqrt y + y\sqrt x\iff \frac14(x+y)\left[2(x+y)+1)\right] \ge x \sqrt y + y\sqrt x$$

which is trivially true for $xy=0$, then assuming $xy\neq 0$ and dividing by $\sqrt{xy}$ it is equivalent to $$\frac14\left(\sqrt{\frac xy}+\sqrt{\frac yx}\right)\left[2(x+y)+1)\right] \ge \sqrt x + \sqrt y$$

which is true indeed since, by AM-GM or rearrangement, $\sqrt{\frac xy}+\sqrt{\frac yx}\ge 2$ we have

$$\frac14\left(\sqrt{\frac xy}+\sqrt{\frac yx}\right)\left(2(x+y)+1)\right) \ge x+y+\frac12\ge \sqrt x + \sqrt y $$

and the latter is true since

$$x+\frac14\ge 2\sqrt{\frac x 4}=\sqrt x$$