# Is this a legal way to prove an inequality?

I have to prove the following inequality:

$(x+y)\sqrt{\frac{x+y}{2}}\geq x\sqrt{y}+y\sqrt{x}$ where $x,y>0$.

After squaring both sides I obtain:

$(x^2+2xy+y^2)\frac{(x+y)}{2}\geq x^2y+xy^2$

then I simplify to

$x^2+y^2\geq 0$.

But this is always true. So my question is does this prove the inequality and if yes why equality is never achieved?!

• Be careful: on the right-hand side, when squaring, $(a + b)^2 \neq a^2 + b^2$. That is $(x\sqrt y + y\sqrt x)^2 \neq x^2 y + y^2 x$. – Namaste Feb 27 '15 at 19:02
• But since both sides are strictly positive numbers then we know that if $a$ and $b$ are positive $\sqrt{a}>\sqrt{b}$ is equivalent to $a>b$. – parkhyeyoo Feb 27 '15 at 19:05
• @parkhyeyoo That doesn't address amWhy's point. – Git Gud Feb 27 '15 at 19:06
• If you really mean legal, then it depends if you have a good lawyer. I guess you mean correct ;-). Try to use clearer English in further questions so that they are understandable easily. – Karl Feb 27 '15 at 19:17
• There are two words that need to be clarified here, namely $obtain$ and $simplify$. Are you using these words to replace $\implies$ or $\iff$? This makes a tremendous difference. – Git Gud Feb 27 '15 at 19:18

Summary answer: No, it is not a proof.

It may be a good way to find a proof, but after you do this, you need to check that the steps work backward.

Sometimes they don't work backward. Example:

PROVE $-2 > -1$: Square both sides, get $4>1$, which is true, done. ???

• But both sides are positive. Then after squaring the resulting inequality is equivalent to the initial..! – parkhyeyoo Feb 27 '15 at 19:24
• Yes, noting both sides are positive (so that taking square-root of both sides is OK) is what you need to do to reverse your first step. But you need to say it. Or, when you are writing the final proof, just do the whole thing in the reverse order. – GEdgar Feb 27 '15 at 19:28
• I think this is not a good example. Because when squaring both sides we get $4<1$. @GEdgar – Fermat Feb 27 '15 at 19:37
• Example. prove that 2=6. Let 2=6 then by the symmetry of the equality we obtain 6=2. By adding two equalities we get 8=8. The proof is complete!!! – Fermat Feb 27 '15 at 19:46

Well here's I think a better proof. Use AM-GM inequality for the square root part and you have:

$$(x+y)\sqrt{\frac{x+y}2} \ge (x+y)\sqrt{xy}$$

Now if we prove that:

$$(x+y)\sqrt{xy} = x^{\frac54}y^{\frac 14} + y^{\frac 54}x^{\frac 14}\ge x\sqrt{y} + y\sqrt{x}= xy^{\frac 12} + yx^{\frac 12}$$

we're done. This follows from Muirhead's inequality, and here's the proof. Divide both sides $x^{\frac 14}y^{\frac 14}$ and we have:

$$x + y \ge x^{\frac34}y^{\frac14} + y^{\frac34}y^{\frac14}$$

Which follows from adding the AM-GM inequalities:

$$\frac{x+x+x+y}{4} \ge \sqrt{x^3y} \quad \text {and} \quad \frac{x+y+y+y}{4} \ge \sqrt{xy^3}$$

Additionally you can prove that the inqeuality holds when $x=y$ using the AM-GM properties

• I know this proof and also a shorter one but the question is whether squaring both sides proves the inequality? – parkhyeyoo Feb 27 '15 at 19:16

Equality is achieved for $x = y$.

Basically this is a valid method for developing the idea for a proof, but not necessarily for actually proving the inequality: you've 'followed your nose' from the inequality you want to prove to an inequality that's easier to understand. What you then need to check is that all those steps work in reverse as well (i.e. that you can start with an inequality that you know to be true, and can work forward to show that implies the desired inequality).

Equivalently, if you work backward through the proof like this, you can just make sure that all the implications are "if and only if", and then there's no need to 'redo' things.^*

In this case, you've made mistakes in your calculations: for instance $(x\sqrt y + y \sqrt x)^2 \neq x^2y + xy^2$ as you state. Further,

$$(x^2 + 2xy + y^2)\dfrac{1}{2}(x + y) = \frac12 x^3 + \frac32xy^2 + \frac32 x^2 y + \frac12y^3$$

*: an example, for real numbers $a,b$

• $a \leq b \Leftrightarrow e^a \leq e^b$

• $a \leq b \leq c \Rightarrow a\leq c$ but $a \leq c \not\Rightarrow a \leq b \leq c$

Basically, when you're using inequalities to imply other inequalities, take some care in thinking about whether you are dealing with equivalences or not.

• Your second paragraph is only correct if you assume the OP isn't writing equivalences but uni-directional implications only. I interpreted it as equivalences. – Git Gud Feb 27 '15 at 19:05
• Sure, that's worth elaborating on. I've clarified this (I'm not convinced OP knows they are writing equivalences). – BaronVT Feb 27 '15 at 19:08