Here is the paper of RMO 2016 Delhi.Pls post answers and solutions to all questions. Also tell ur marks and estimated cut off Thanks!

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TopNewestProblem 4

\(f\left( a,b,c \right) =\quad \sum { \frac { a }{ { a }^{ 3 }+{ b }^{ 2 }+{ c } } } =\quad \sum { \frac { 1 }{ { a }^{ 2 }+\frac { { b }^{ 2 } }{ a } +\frac { c }{ a } } } \\ \\ Now,\quad \frac { { a }^{ 2 }+{ \frac { { b }^{ 2 } }{ a } + }{ \frac { c }{ a } } }{ 3 } \ge \sqrt [ 3 ]{ { b }^{ 2 }{ c } } \quad \Rightarrow \quad \frac { 1 }{ { a }^{ 2 }+\frac { { b }^{ 2 } }{ a } +\frac { c }{ a } } \le \quad \frac { 1 }{ 3\sqrt [ 3 ]{ { b }^{ 2 }c } } \left\{ A.M-G.M \right\} \\ \sum { \frac { 1 }{ { a }^{ 2 }+\frac { { b }^{ 2 } }{ a } +\frac { c }{ a } } } \le \quad \sum { \frac { 1 }{ 3\sqrt [ 3 ]{ { b }^{ 2 }c } } } \le \sqrt [ 3 ]{ abc } \le \frac { a+b+c }{ 3 } =1\quad \\ \\ \Rightarrow f\left( a,b,c \right) \le 1,\quad with\quad equality\quad at\quad a=b=c=1\) – Aditya Dhawan · 5 months, 2 weeks ago

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– Kaustubh Miglani · 5 months, 2 weeks ago

Will my answer be correct?Log in to reply

– Kaustubh Miglani · 5 months, 2 weeks ago

I USED Titus lemma. Got same answerLog in to reply

– Yuvraj Singh · 2 months, 3 weeks ago

THEN U GOTTA PROVE IT COZ IT'S NOT IN STANDARD TEXT BOOKSLog in to reply

– Kaustubh Miglani · 2 months, 2 weeks ago

It is in standard books.Atleast cauchy isLog in to reply

Problem 1

We can see that as we move the point \(P\) on the circumference of the circle\([\)excluding \(X\) and \(Y],\)the \(\angle XPY=\angle XP_1Y\) remains constant.So this shows that \(AB=A_1B_1.\)Now we use extended sin rule to complete the problem.

Let the circum-radius of \(\triangle PAB\) be \(R\) and \(\triangle P_1A_1B_1\) be \(R_1.\)

In \(\triangle PAB, \frac{AB}{sin\angle P}=2R\) and in \(\triangle P_1A_1B_1,\frac{A_1B_1}{sin\angle P_1}=\frac{AB}{sin\angle P}=2R_1.\) Therefore \(2R=2R_1\Rightarrow\boxed {R=R_1}.\)Hence Proved. – Ayush Rai · 5 months, 2 weeks ago

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– Abhishek Alva · 5 months, 2 weeks ago

nice solutionLog in to reply

– Kaustubh Miglani · 2 months, 3 weeks ago

Hey can u explain in detail why A1B1=ABLog in to reply

– Yuvraj Singh · 2 months, 3 weeks ago

Because angle subtended by both of these chords at centre are equal. angle A1 X A =P X P1=P Y P1=B Y B1Log in to reply

– Kaustubh Miglani · 2 months, 2 weeks ago

Which class are u in and which school?Log in to reply

– Kaustubh Miglani · 2 months, 2 weeks ago

I m in class 10 And from bbps dw. But ur profile says u live in noidaLog in to reply

– Yuvraj Singh · 2 months, 2 weeks ago

I am graduating from IIT Bombay. What is ur favourite college bro?Log in to reply

– Kaustubh Miglani · 2 months, 2 weeks ago

IIT Is the one Bombay ,PerhapsLog in to reply

– Yuvraj Singh · 2 months, 2 weeks ago

Prefered Jee rank?mine was 17Log in to reply