# Thread: 2 Inequality Proofs

1. ## 2 Inequality Proofs

1. Prove that $\displaystyle \frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\geq a+b+c$ given that a, b, and c are positive real numbers.

----------------------------

2. Prove that $\displaystyle \frac{x^3}{yz}+\frac{y^3}{zx}+\frac{z^3}{xy}\geq x+y+z$ given that x, y, and z are positive real numbers.

2. Originally Posted by Winding Function
1. Prove that $\displaystyle \frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\geq a+b+c$ given that a, b, and c are positive real numbers.

----------------------------

2. Prove that $\displaystyle \frac{x^3}{yz}+\frac{y^3}{zx}+\frac{z^3}{xy}\geq x+y+z$ given that x, y, and z are positive real numbers.
the questions you've been asking do not belong to "Advanced Algebra" forum. we have a forum for inequalities. anyway, here's a solution to your problems:

expanding $\displaystyle (u-v)^2 + (v - w)^2 + (w-u)^2 \geq 0$ gives us: $\displaystyle u^2 + v^2 + w^2 \geq uv + vw + wu. \ \ \ \ (1)$

therefore $\displaystyle u^4 + v^4 + w^4 \geq (uv)^2 + (vw)^2 + (wu)^2.$ but again by (1): $\displaystyle (uv)^2 + (vw)^2 + (wu)^2 \geq uvw(u+v+w).$ thus: $\displaystyle u^4 + v^4 + w^4 \geq uvw(u+v+w). \ \ \ \ (2)$

now in (1) put $\displaystyle u=ab, \ v =bc, \ w=ac$ and then divide by $\displaystyle abc$ to prove your first problem and in (2) put $\displaystyle u=x, \ v = y, \ w=z$ and then divide by $\displaystyle xyz$ to prove your second problem. $\displaystyle \Box$

3. 1.

By Cauchy-Schwarz: $\displaystyle \left( {\tfrac{{bc}} {a} + \tfrac{{ac}} {b} + \tfrac{{ab}} {c}} \right) \cdot \left( {\tfrac{{ab}} {c} + \tfrac{{bc}} {a} + \tfrac{{ac}} {b}} \right) \geqslant \left( {a + b + c} \right)^2$ - since $\displaystyle \left( {\tfrac{{bc}} {a} \cdot \tfrac{{ab}} {c} = b^2 ;...} \right)$ -

Thus: $\displaystyle \left( {\tfrac{{bc}} {a} + \tfrac{{ac}} {b} + \tfrac{{ab}} {c}} \right)^2 \geqslant \left( {a + b + c} \right)^2$ and by the positivity of both sums we have: $\displaystyle \tfrac{{bc}} {a} + \tfrac{{ac}} {b} + \tfrac{{ab}} {c} \geqslant a + b + c$

2.

I'll use the lemma here. (1)

Using it we have: $\displaystyle \left( {\tfrac{{x^3 }} {{yz}} + \tfrac{{y^3 }} {{xz}} + \tfrac{{z^3 }} {{xy}}} \right) \cdot \left( {yz + xz + xy} \right) \cdot \left( {1 + 1 + 1} \right) \geqslant \left( {x + y + z} \right)^3$

Now since $\displaystyle x^2 + y^2 + z^2 \geqslant yz + xz + xy$ see in the link in (1), we have $\displaystyle x^2 + y^2 + z^2 + 2 \cdot \left( {yz + xz + xy} \right) \geqslant 3 \cdot \left( {yz + xz + xy} \right)$ i.e. $\displaystyle \left( {x + y + z} \right)^2 \geqslant 3 \cdot \left( {yz + xz + xy} \right)$

Thus we have: $\displaystyle \left( {\tfrac{{x^3 }} {{yz}} + \tfrac{{y^3 }} {{xz}} + \tfrac{{z^3 }} {{xy}}} \right) \cdot 3 \geqslant \tfrac{{\left( {x + y + z} \right)^2 }} {{yz + xz + xy}} \cdot \left( {x + y + z} \right) \geqslant 3 \cdot \left( {x + y + z} \right)$ $\displaystyle \Rightarrow \tfrac{{x^3 }} {{yz}} + \tfrac{{y^3 }} {{xz}} + \tfrac{{z^3 }} {{xy}} \geqslant x + y + z \square$

4. On the first inequality, we can start by using
$\displaystyle x^2(y-z)^2+y^2(z-x)^2+z^2(x-y)^2=2(x^2y^2+y^2z^2+x^2z^2-x^2yz-xy^2z-xyz^2$
(try expanding the left hand side to see that this is true).
Thus
$\displaystyle \frac{x^2(y-z)^2+y^2(z-x)^2+z^2(x-y)^2}{2xyz}=\frac{x^2y^2+y^2z^2+x^2z^2-x^2yz-xy^2z-xyz^2}{xyz}$
$\displaystyle =\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}-x-y-z$
But if x,y, and z are all positive, then xyz is positive. And as $\displaystyle x^2(y-z)^2+y^2(z-x)^2+z^2(x-y)^2$ is the sum of three squares, it cannot be negative. Thus $\displaystyle \frac{x^2(y-z)^2+y^2(z-x)^2+z^2(x-y)^2}{2xyz}\ge0$. From our above equality, then, we have
$\displaystyle \frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}-x-y-z\ge0$
$\displaystyle \frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}\ge{x+y+z}$

--Kevin C.

5. Originally Posted by Winding Function
1. Prove that $\displaystyle \frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\geq a+b+c$ given that a, b, and c are positive real numbers.
That should be simple enough. Let's see....

$\displaystyle \frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\geq a+b+c$

$\displaystyle a^2b^2+a^2c^2+b^2c^2\geq a^2bc+ab^2c+abc^2$

By symmetry, let $\displaystyle c\geq b\geq a$. Then:

$\displaystyle a^2c^2+b^2c^2-2abc^2\geq a^2bc+ab^2c-a^2b^2-abc^2$

$\displaystyle c^2(a^2+b^2-2ab)\geq ab(ac+bc-ab-c^2)$

$\displaystyle c^2(b-a)^2\geq ab(c-b)(a-c)$

We can see that $\displaystyle c^2(b-a)^2\geq 0$, whereas $\displaystyle ab(c-b)(a-c)\leq0$ (because $\displaystyle a-c\leq 0$).

----------------------------

2. Prove that $\displaystyle \frac{x^3}{yz}+\frac{y^3}{zx}+\frac{z^3}{xy}\geq x+y+z$ given that x, y, and z are positive real numbers.

6. Originally Posted by NonCommAlg
the questions you've been asking do not belong to "Advanced Algebra" forum. we have a forum for inequalities.
Where is the forum for inequalities? I don't think these problems fit under the category of "Solving and Graphing Inequalities" for Elementary and Middle School Math Help (http://www.mathhelpforum.com/math-help/inequalities/).

Also, I don't understand your solution. Could you please indicate which parts correspond to which question? Thanks!