for all positive integers \(n \ge 2\)
Solution:
Using the AM-HM Inequality we have
The RHS on simplification becomes \(\frac{n^2}{n-1}\). This is known as the famous Nesbitt’s Inequality
Q-2) Find the maximum value of \((7-x)^4(2+x)^5\) with \(-2\le x \le 7\).
Solution: Set \(p=7-x\) and \(q=2+x\). We thus have \(p+q=9\), and we need the maximum value of \(p^4q^5\).
Applying the AM-GM we have
This gives us
Q-3) If \(2x^3+ax^2+bx+4=0\) (\(a,b\) are positive reals) has 3 real roots, then show that \(a+b \ge 6(2^{\frac{1}{3}}+4^{\frac{1}{3}})\)
Solution :
Let \(\alpha, \beta, \gamma\) be the 3 real roots of the equation.
Gievn that all the coefficients are positive, we know that all the roots are thus negative (Descartes!). So let’s have the substitution \(\alpha \rightarrow -\alpha, \beta \rightarrow -\beta, \gamma \rightarrow -\gamma\) with all \(\alpha, \beta, \gamma \) positive.
We have \(\sum \alpha = \frac{a}{2}\), \(\sum \alpha\beta = \frac{b}{2}\) and \(\sum \alpha\beta\gamma = 2\). AM-GM gives
\(\frac{\alpha + \beta + \gamma}{3} \ge (\alpha \beta \gamma)^{\frac{1}{3}}\Rightarrow \frac{a^3}{216} \ge 2\Rightarrow a\ge 6\times 2^{\frac{1}{3}}\).
Similarly AM-GM on \(\sum \alpha\beta\) gives \(\frac{b^3}{216}\ge 4\Rightarrow b \ge 6\times 4^{\frac{1}{3}}\). Adding these 2 results we have our proof.
Q-4) Show that \((n!)^{\frac{1}{n}} < \frac{n+1}{2}\)
Solution :
Consider AM-GM on the first \(n\) numbers. We have
\(\frac{1+2+3+4…+n}{n}>(1\times 2\times 3…\times n)^{\frac{1}{n}}\Rightarrow \frac{n+1}{2} >(n!)^{\frac{1}{n}} \)..
Q-5) If \(a,b,c,d\) are positive reals such that \(a+b+c+d=2\) and \(M=(a+b)(c+d)\), then show that \(M \le 1\)
Solution:
Assuming that \(M \ge 1\), we have \(a+b \ge \frac{1}{c+d}\Rightarrow a+b+c+d \ge c+d +\frac{1}{c+d}\Rightarrow 2 \ge c+d+\frac{1}{c+d}\), which is a contradiction.
Q-6) Let the equation \(x^2-3x+p=0\) have 2 positive roots \(a,b\). Find the minimum value of \(\left ( \frac{4}{a}+\frac{1}{b}\right)\).
Solution:
We have \(a+b=3\), applying AM-HM on the numbers \(\frac{a}{2}, \frac{a}{2}, b\) we have
Q-7) For positive reals \(a,b,c\) with \(abc=1\) show that \(a^{b+c}b^{c+a}c^{a+b} \le 1\)
Solution:
Withput loss of generality we can assume \(a \le b \le c\), we can make this assumption since the expression \(a^{b+c}b^{c+a}c^{a+b}\) is symmetric in \(a,b,c\).
We thus have \(a \le 1 \le c\). Also \(c-b \ge 0\) & \(b-a \ge 0\).
This gives us
The required expression we have is :
Q-8) If \(x,y\in\mathbb{R}^+\), with \(m,n \in \mathbb{N}\), show that \(\frac{x^ny^m}{(1+x^{2n})(1+y^{2m})} \le \frac{1}{4}\)
Solution :
We have
and
Multiplying the 2 we have
An alternate approach may have been by applying the well-known inequality \(x+\frac{1}{x} \ge 2\) for \(x=x^n\) & \(x=y^m\).
Q-9) If \(a\in \mathbb{R}\) is a root of the equation \(x^5-x^3+x-2=0\), show that \(\left \lfloor a^6 \right \rfloor =3\). (You may need to read up on the Floor function)
Solution:
Note that from the given equation we have \(a[(a^2-1)^2+a^2]=2\Rightarrow a>0\).
We have from AM-GM \(a^3+2=a^5+a>2a^3\Rightarrow a^6<4\)
Which gives \(\left \lfloor a^6 \right \rfloor =3\).
Q-10) Let \(x>0,y>0\) be reals satisfying \(x+y=2\). Show that
a) \(x^2y^2(x^2+y^2) \le 2\)
b) \(x^3y^3(x^3+y^3) \le 2\)
Solution:
a) Put \(x=1-t\) and \(y=1+t\).
We thus have
\(x^2y^2(x^2+y^2)=2(1-t^2)(1-t^4) \le 2\).
b) Using \(x^3+y^3=(x+y)(x^2-xy+y^2)=2(4-3xy)\), we have on applying of AM-GM on \((4-3xy),xy,xy,xy\)
Thus we have
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