# proofs question (1 Viewer)

#### cormglakes

##### New Member

so i can do question A by doing LHS- RHS and then doing 2 cases where a>b and a<b so show LHS - RHS >0 ; is there and alternative solution?

and for B and C, I have absolutely no clue.; for B i think you have to add multiple inequalities but i am not sure what inequalities.

#### cormglakes

##### New Member
Also another proofs question,
how the bloody hell do u prove :

a/(b+c) + b/(a+c) + c/(a+b) > 3/2

also does anyone have proofs tips? coz i rlly suck at proofs.

#### idkkdi

##### Well-Known Member
Also another proofs question,
how the bloody hell do u prove :

a/(b+c) + b/(a+c) + c/(a+b) > 3/2

also does anyone have proofs tips? coz i rlly suck at proofs.
View attachment 30433
so i can do question A by doing LHS- RHS and then doing 2 cases where a>b and a<b so show LHS - RHS >0 ; is there and alternative solution?

and for B and C, I have absolutely no clue.; for B i think you have to add multiple inequalities but i am not sure what inequalities.
These one's are pretty hard. I've done them before but I seriously can't be bothered to write it.
A few ways for both.
1. Amgm with two or three terms can get the job done. Imaginary terms should be created. Referring to part C.
2. Am-gm with three terms of a two term method also exists. Start from something else to get to this answer, backwards reasoning may help but experience likely needed.

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#### CM_Tutor

##### Moderator
Moderator
Theorem 1: For all positive integers $\bg_white a$, $\bg_white b$, and $\bg_white c$,

$\bg_white a^3 + b^3 \geqslant abc\left(\frac{a}{c}+\frac{b}{c}\right)$

Assume the theorem is false. Then there exist positive integers $\bg_white a$, $\bg_white b$, and $\bg_white c$, such that

\bg_white \begin{align*} a^3 + b^3 &< abc\left(\frac{a}{c}+\frac{b}{c}\right) \\ a^3 + b^3 &< \frac{a^2bc}{c}+\frac{ab^2c}{c} \\ a^3 + b^3 &< a^2b +ab^2 \\ a^3 + b^3 &< ab(a +b) \\ \frac{a^3 + b^3}{a+b} &< ab \\ \frac{(a+b)(a^2 - ab + b^2)}{a+b} &< ab \\ a^2 - ab + b^2 &< ab \\ a^2 - 2ab + b^2 &< 0 \\ (a - b)^2 &< 0 \end{align*}

This is a contradiction as $\bg_white a - b$ is an integer and so its square cannot be negative.

Every statement following the assumption is true, provided the assumption is true - and yet, the result is a statement that is false. This is only possible if the assumption itself is false, and thus the theorem must be true.

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#### CM_Tutor

##### Moderator
Moderator
Theorem 1: For all positive integers $\bg_white a$, $\bg_white b$, and $\bg_white c$,

$\bg_white a^3 + b^3 \geqslant abc\left(\frac{a}{c}+\frac{b}{c}\right)$

Direct Proof:

\bg_white \begin{align*} \text{Since a and b are integers} \qquad (a - b)^2 &\geqslant 0 \\ a^2 - 2ab + b^2 &\geqslant 0 \\ a^2 - ab + b^2 &\geqslant ab \\ \frac{(a+b)(a^2 - ab + b^2)}{a+b} &\geqslant ab \qquad \text{as a + b > 0} \\ \frac{a^3 + b^3}{a+b} &\geqslant ab \\ a^3 + b^3 &\geqslant ab(a +b) \\ a^3 + b^3 &\geqslant \frac{abc}{c}(a +b) \qquad \text{as c > 0} \\ a^3 + b^3 &\geqslant abc\left(\frac{a}{c} + \frac{b}{c}\right) \end{align*}

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#### CM_Tutor

##### Moderator
Moderator
Theorem 2: For all positive integers $\bg_white a$, $\bg_white b$, and $\bg_white c$,

$\bg_white \frac{a^3 + b^3 + c^3}{3} \geqslant abc$

Direct Proof:

From Theorem 1, we know that

$\bg_white a^3 + b^3 \geqslant abc\left(\frac{a}{c} + \frac{b}{c}\right) \qquad \qquad \text{. . . . . . . . . (1)}$

And by changing the dummy variables, we also know that

$\bg_white b^3 + c^3 \geqslant abc\left(\frac{b}{a} + \frac{c}{a}\right) \qquad \qquad \text{. . . . . . . . . (2)}$

and that

$\bg_white a^3 + c^3 \geqslant abc\left(\frac{a}{b} + \frac{c}{b}\right) \qquad \qquad \text{. . . . . . . . . (3)}$

Adding (1) + (2) + (3) gives:

\bg_white \begin{align*} 2\left(a^3 + b^3 + c^3\right) &\geqslant abc\left(\frac{a}{c} + \frac{b}{c} + \frac{b}{a} + \frac{c}{a} + \frac{a}{b} + \frac{c}{b}\right) \\ &\geqslant abc\left(\frac{a}{c} + \frac{c}{a} + \frac{b}{a} + \frac{a}{b} + \frac{b}{c} + \frac{c}{b}\right) \qquad \qquad \text{. . . . . . . . . (*)} \end{align*}

Notice that we have, on the RHS of (*), three pairs of terms all of the form

$\bg_white \frac{x}{y} + \frac{y}{x}$

where $\bg_white x$ and $\bg_white y$ are positive integers. This can be simplified by recognising that each of these pairs of terms has a minimum value of 2:

\bg_white \begin{align*} \left(\sqrt{\frac{x}{y}} - \sqrt{\frac{y}{x}}\right)^2 &\geqslant 0 \\ \frac{x}{y} - 2\sqrt{\frac{x}{y}\times\frac{y}{x}} + \frac{y}{x} &\geqslant 0 \\ \frac{x}{y} - 2 + \frac{y}{x} &\geqslant 0 \\ \frac{x}{y} + \frac{y}{x} &\geqslant 2 \end{align*}

Applying this result to (*) gives:

\bg_white \begin{align*} 2\left(a^3 + b^3 + c^3\right) &\geqslant abc\left[\left(\frac{a}{c} + \frac{c}{a}\right) + \left(\frac{b}{a} + \frac{a}{b}\right) + \left(\frac{b}{c} + \frac{c}{b}\right)\right] \geqslant abc\left(2 + 2 + 2\right) \\ 2\left(a^3 + b^3 + c^3\right) &\geqslant 6abc \\ \frac{2}{6}\left(a^3 + b^3 + c^3\right) &\geqslant abc \\ \frac{a^3 + b^3 + c^3}{3} &\geqslant abc \end{align*}

#### CM_Tutor

##### Moderator
Moderator
Theorem 3: For all positive integers $\bg_white a$, $\bg_white b$, and $\bg_white c$,

$\bg_white \frac{a^3}{b} + \frac{b^3}{c} + \frac{c^3}{a} \geqslant ab + bc + ca$

Direct Proof:

In the direct proof of Theorem 1, it was established that

$\bg_white a^3 + b^3 \geqslant ab(a +b)$

from which it can be shown (by division by $\bg_white b$) that:

\bg_white \begin{align*} \frac{a^3}{b} + \frac{b^3}{b} &\geqslant \frac{ab}{b}(a +b) \\ \frac{a^3}{b} + b^2 &\geqslant a^2 + ab \qquad \qquad \text{. . . . . . . . . (1)} \end{align*}

Using the same approach with $\bg_white b$ and $\bg_white c$, we find that:

$\bg_white \frac{b^3}{c} + c^2 \geqslant b^2 + bc \qquad \qquad \text{. . . . . . . . . (2)}$

and with $\bg_white a$ and $\bg_white c$, we get

$\bg_white \frac{c^3}{a} + a^2 \geqslant c^2 + ac \qquad \qquad \text{. . . . . . . . . (3)}$

\bg_white \begin{align*} \left(\frac{a^3}{b} + b^2\right) + \left(\frac{b^3}{c} + c^2\right) + \left(\frac{c^3}{a} + a^2\right) &\geqslant \left(a^2 + ab\right) + \left(b^2 + bc\right) + \left(c^2 + ac\right) \\ \frac{a^3}{b} + \frac{b^3}{c} + \frac{c^3}{a} + a^2 + b^2 + c^2 &\geqslant ab + bc + ca + a^2 + b^2 + c^2 \\ \frac{a^3}{b} + \frac{b^3}{c} + \frac{c^3}{a} &\geqslant ab + bc + ca \end{align*}

#### TheShy

##### Active Member
a/(b+c) + b/(a+c) + c/(a+b) > 3/2

You can consider doing this by considering the fact that: (a+b+c)(1/a + 1/b + 1/c)>=9. Of course you have to prove that first, but its very simple, just expand and use the fact that a+1/a >=2.
Anyways, considering that, let a=b+c, b=a+c, c=a+b. Sub those in, then expand, and you get the desired result

#### slightly_above_average

##### New Member
for part a)
Expand right hand side
bring terms to the left
factorise by (a-b)
so u get (a^2-b^2)(a-b)
so youll get (a+b)(a-b)^2 which should right

#### s97127

##### Member
a^3+b^3+c^3 - 3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca) = (a+b+c) * 1/2 * [(a-b)^2 + (b-c)^2 + (c-a)^2] >= 0

This one is stronger as we don't need a, b and c are positive numbers. We only need a+b+c >= 0

#### s97127

##### Member
a/(b+c) + b/(a+c) + c/(a+b) > 3/2

let x = a+b, y = a+c and z = a+b
the problem will become:
1/2 * [(y+z-x)/x + (x+z-y)/y + (x+y-z)/z) >= 3/2
(y/x + x/y) + (z/x + x/z) + (y/z + z/y) >= 6
this is easy as y/x + x/y >= 2