Each question is worth 7 points

Time allowed is 4 hours

No books, notes or calculators permitted

Write full proofs with your answers

**1)** Find all real numbers \(a\) such that there are four different real numbers \(x\) satisfying the equation

\[x(x+1)^3=(2x+a)(x+a-1).\]

**2)** Find all functions \(f:\mathbb{R} \rightarrow \mathbb{R}\) such that for \(x, y \in \mathbb{R}\) we have

\[f(x f(y) + x) = xy+f(x).\]

**3)** Suppose that \(x\), \(y\) and \(z\) are real numbers satisfying the following three equations.

\[\begin{align} x+y+z &=2\\ x^2 +y^2 + z^2 &=6\\ x^3+y^3+z^3&=8 \end{align}\]

Find all possible values of \(x^{2015} + y^{2015} + z^{2015}\).

**4)** Suppose that a function \(f:\mathbb{R} \rightarrow \mathbb{R}\) has the property that

\[f(x)^2 \leq f(y)\]

for every \(x > y\).

Prove that \(0 \leq f(x) \leq 1\) for all \(x \in \mathbb{R}\).

**5)** Prove the following inequality for all \(a, b, c > 0\).

\[\dfrac {a^2(a^3 + b^3)}{a^2 + ab + b^2} +\dfrac {b^2(b^3 + c^3)}{b^2 + bc + c^2} +\dfrac {c^2(c^3 + a^3)}{c^2 + ca + a^2} \geq 2abc\]

## Comments

Sort by:

TopNewest3). let \(s_1,s_2,s_3\) be the symmetric sums. let \(p_n\) denote the nth power sum. then by newtons sum: \[\begin{array}a p_1=s_1=2\\ p_2=s_1p_1-2s_2=4-2s_2=6\Longrightarrow s_2=-1\\ p_3=s_1p_2-s_2p_1+3s_3=14+3s_3=8\Longrightarrow s_3=-2\end{array}\] so, x,y,z are roots of \[x^3-2x^2-x+2=0\] \[x^3-x-2x^2+2=x(x^2-1)-2(x^2-1)=(x-2)(x-1)(x+1)=0\] since the expression is symmetric, order doesnt matter. we have \((x,y,z)=(2,1,-1)\). put this in to get \[x^{2015}+y^{2015}+z^{2015}=\boxed{2^{2015}}\] this is the only value possible. are we suppose to expand \(2^{2015}\)? – Aareyan Manzoor · 1 year, 3 months ago

Log in to reply

– Sharky Kesa · 1 year, 3 months ago

Do you think we do? Of course not! Nice solution. 7 out of 7.Log in to reply

– Seong Ro · 1 year, 3 months ago

Is the answer to the second question is \(f(x)=0\)Log in to reply

– Sharky Kesa · 1 year, 3 months ago

No. Also, please give proof.Log in to reply

\(\large{a \in \left(\frac{1}{24} (27-\sqrt{3}),\frac{1}{24}(27+\sqrt{3}) \right)}\) – Seong Ro · 1 year, 3 months ago

Log in to reply

– Sharky Kesa · 1 year, 3 months ago

No. Also, please give proof with your answers.Log in to reply

\(\large{a \in \left(\frac{1}{432} (153-29 \sqrt{57}),\frac{1}{432}(153+29 \sqrt{57}) \right)}\) – Seong Ro · 1 year, 3 months ago

Log in to reply

For roots to be real and distinct its derivative must have three distinct roots

\(4x^3+6x^2+2x+3(a-1)\).....\((1)\) And for that to happen , we again take its derivative, we get

\(12x^2+12x+2=0\)

For real and distinct roots

\(f(x_{1}) f(x_{2})<0\) where \(x_{1,2}=\frac{-3 \pm \sqrt{3}}{4}\) in \((1)\)

and then i found interval for \(a\) – Seong Ro · 1 year, 3 months ago

Log in to reply

– Seong Ro · 1 year, 3 months ago

As my answers are wrong , proof would be useless :(Log in to reply

@Sharky Kesa : For question 1 my solution is as follows : The given equation simplifies to x^4+3.x^3+x^2+(3-3a)x +(a-a^2)=0. Let the roots be p,q,r,s. Now, sum of roots two at a time=1, sum of roots three at a time =(3a-3) and all four roots at a time=(a-a^2)

Now we use the identity : (pqr+pqs+prs+qrs)^2=(p^2.q^2.r^2 + .........) + 2pqrs(pq+pr+ps+qr+qs+rs) Thus we obtain the inequality : (pqr+pqs+prs+qrs)^2 >= 3pqrs(pq+pr+ps+qr+qs+rs) Substituting the values we obtain a condition : 4a^2-7a+3 >=0. Solving that we obtain that a should lie outside [3/4,1].

Please tell me whether the answer and the solution is correct or wrong. – Shrihari B · 1 year, 2 months ago

Log in to reply

Could you provide the solutions to all questions please? So that we can reference and improve. Thanks! – Lai Hong Chiu · 1 year, 3 months ago

Log in to reply

Using \(AM-GM\) \[a^{2}+b^{2}\geq2ab\] \[\frac{3(a^{2}+b^{2})}{2}\geq(a^{2}+ab+b^{2})\] This implies \[\frac{a^{2}(a^{3}+b^{3})}{a^{2}+ab+b^{2}}\geq\frac{2a^{2}(a^{3}+b^{3})}{3(a^{2}+b^{2})}\] Using Generalized Power mean inequality We get \[\frac{a^{2}(a^{3}+b^{3})}{a^{2}+ab+b^{2}}\geq\frac{2a^{2}(\sqrt{\frac{a^{2}+b^{2}}{2}})}{3}\] Now by \(RMS-AM\) inequality we get \[\frac{a^{2}(a^{3}+b^{3})}{a^{2}+ab+b^{2}}\geq\frac{a^{2}(a+b)}{3}\geq\frac{2a^{2}\sqrt{ab}}{3}\] Therefore \[\sum_{cyc}^{a,b,c}\frac{a^{2}(a^{3}+b^{3})}{a^{2}+ab+b^{2}}\geq\sum_{cyc}^{a,b,c}\frac{2a^{2}\sqrt{ab}}{3}\] Now using simple \(AM-GM\) on the RHS of the above inequality we get \[\sum_{cyc}^{a,b,c}\frac{a^{2}(a^{3}+b^{3})}{a^{2}+ab+b^{2}}\geq\sum_{cyc}^{a,b,c}\frac{2a^{2}\sqrt{ab}}{3}\geq2abc\]

\[\sum_{cyc}^{a,b,c}\frac{a^{2}(a^{3}+b^{3})}{a^{2}+ab+b^{2}}\geq2abc\] Hence proved . – Shivam Jadhav · 1 year, 3 months ago

Log in to reply

Putting \(y=0\) we get \[f(xf(0)+x)=f(x))\] \[xf(0)+x=x\] \[\boxed{f(0)=0}.........(1)\] Now we assume that there exists \(f(y)=-1\) Then putting it in original functional equation we get, \[f(0)=xy+f(x)\] \[-xy=f(x)........(2)\] now put \(x=y\) \[f(y)=-y^{2}\] But \(f(y)=-1\) Therefore \(y=1\) or \(y=-1\). Now putting these values in \((2)\) we get \(f(x)=x\) or \(f(x)=-x\) . since above two functions satisfy the given functional equation and \((1)\) Hence they are the solutions of the given functional equation. – Shivam Jadhav · 1 year, 3 months ago

Log in to reply

– Sharky Kesa · 1 year, 3 months ago

You would get 3 out of 7 for this proof. In line 2, how do you know \(f\) is injective?Log in to reply

– Shrihari B · 1 year, 2 months ago

Proving that the function is injective is very easy. For two reals a and b let f(a)=f(b). We take two cases. In the first case put y=a and in second case put y=b. Now note that in both cases the LHS is the same. Therefore comparing the RHS you obtain that a=b. Hence function is injectiveLog in to reply

– Shivam Jadhav · 1 year, 3 months ago

Answer is right I guess (function)Log in to reply

– Sharky Kesa · 1 year, 3 months ago

Yeah it is. But you have done some assumptions.Log in to reply