Algebra : max and min: RMO/INMO problem solving practice

Question 1:

If x^{2}+y^{2}=c^{2}, find the least value of \frac{1}{x^{2}} + \frac{1}{y^{2}}.

Solution 1:

Let z^{'}=\frac{1}{x^{2}} + \frac{1}{y^{2}} = \frac{y^{2}+x^{2}}{x^{2}y^{2}} = \frac{c^{2}}{x^{2}y^{2}}

z^{'} will be minimum when \frac{x^{2}y^{2}}{c^{2}} will be minimum.

Now, let z=\frac{x^{2}y^{2}}{c^{2}}=\frac{1}{c^{2}}(x^{2})(y^{2})….call this equation I.

Hence, z will be maximum when x^{2}y^{2} is maximum but (x^{2})(y^{2}) is the product of two factors whose sum is x^{2}+y^{2}=c^{2}.

Hence, x^{2}y^{2} will be maximum when both these factors are equal, that is, when

\frac{x^{2}}{1}=\frac{y^{2}}{1}=\frac{x^{2}+y^{2}}{1}=\frac{c^{2}}{1}. From equation I, maximum value of z=\frac{c^{2}}{4}. Hence, the least value of \frac{1}{x^{2}} + \frac{1}{y^{2}}=\frac{4}{c^{2}}.

Some basics related to maximum and minimum:

Basic 1:

Let a and b be two positive quantities, S their sum and P their product; then, from the identity:

4ab=(a+b)^{2}-(a-b)^{2}, we have

4P=S^{2}-(a-b)^{2} and S^{2}=4P+(a-b)^{2}.

Hence, if S is given, P is greatest when a=b; and if P is given, S is least when a=b. That is, if the sum of two positive quantities is given, their product is greatest when they are equal; and, if the product of two positive quantities is given, their sum is least when they are equal.

Basic 2:

To find the greatest value of a product the sum of whose factors is constant.

Solution 2:

Let there be n factors a,b,c,\ldots, k, and suppose that their sum is constant and equal to s.

Consider the product abc\ldots k, and suppose that a and b are any two unequal factors. If we replace the two unequal factors a and b by the two equal factors \frac{a+b}{2}, \frac{a+b}{2}, the product is increased, while the sum remains unaltered; hence, so long as the product contains two unequal factors it can be increased without altering the sum of the factors; therefore, the product is greatest when all the factors are equal. In this case, the value of each of the n factors is \frac{s}{m}, and the greatest value of the product is (\frac{s}{n})^{n}, or (\frac{a+b+c+\ldots+k}{n})^{n}.

Corollary to Basic 2:

If a, b, c, \ldots k are unequal, (\frac{a+b+c+\ldots+k}{n})^{2}>abc\ldots k;

that is, \frac{a+b+c+\ldots +k}{n} > (\frac{a+b+c+\ldots + k}{n})^{\frac{1}{n}}.

By an extension of the meaning of the terms arithmetic mean and geometric mean, this result is usually stated as follows: the arithmetic mean of any number of positive quantities is greater than the geometric mean.

Basic 3:

To find the greatest value of a^{m}b^{n}c^{p}\ldots when a+b+c+\ldots is constant; m,n, p, ….being positive integers.

Solution to Basic 3:

Since m,n,p, …are constants, the expression a^{m}b^{n}c^{p}\ldots will be greatest when (\frac{a}{m})^{m}(\frac{b}{n})^{n}(\frac{c}{p})^{p}\ldots is greatest. But, this last expression is the product of m+n+p+\ldots factors whose sum is m(\frac{a}{m})+n(\frac{b}{n})+p(\frac{c}{p})+\ldots, or a+b+c+\ldots, and therefor constant. Hence, a^{m}b^{n}c^{p}\ldots will be greatest when the factors \frac{a}{m}, \frac{b}{n}, \frac{c}{p}, ldots are all equal, that is, when

\frac{a}{m} = \frac{b}{n} = \frac{c}{p} = \ldots = \frac{a+b+c+\ldots}{m+n+p+\ldots}

Thus, the greatest value is m^{m}n^{n}p^{p}\ldots (\frac{a+b+c+\ldots}{m+n+p+\ldots})^{m+n+p+\ldots}.

Some examples using the above techniques:

Example 1:

Show that (1^{r}+2^{r}+3^{r}+\ldots+n^{r})>n^{n}(n!)^{r} where r is any real number.

Solution 1:

Since \frac{1^{r}+2^{r}+3^{r}+\ldots+n^{r}}{n}>(1^{r}.2^{r}.3^{r}\ldots n^{r})^{\frac{1}{n}}

Hence, (\frac{1^{r}+2^{r}+3^{r}+\ldots+n^{r}}{n})^{n}>1^{r}.2^{r}.3^{r} \ldots n^{r}

that is, >(n!)^{r}, which is the desired result.

Example 2:

Find the greatest value of (a+x)^{3}(a-x)^{4} for any real value of x numerically less than a.

Solution 2:

The given expression is greatest when (\frac{a+x}{3})^{3}(\frac{a-x}{4})^{4} is greatest; but, the sum of the factors of this expression is 3(\frac{a+x}{3})+4(\frac{a-x}{4}), that is, 2a; hence, (a+x)^{3}(a-x)^{4} is greatest when \frac{a+x}{3}=\frac{a-x}{4}, that is, x=-\frac{a}{7}. Thus, the greatest value is \frac{6^{3}8^{4}}{7^{7}}a^{r}.

Some remarks/observations:

The determination of maximum and minimum values may often be more simply effected by the solution of a quadratic equation than by the foregoing methods. For example:

Question:

Divide an odd integer into two integral parts whose product is a maximum.

Answer:

Let an odd integer be represented as 2n+1; the two parts by x and 2n+1-x; and the product by y; then (2n+1)x-x^{2}=y; hence,

2x=(2n+1)\pm \sqrt{(2n+1)^{2}-4y}

but the quantity under the radical sign must be positive, and therefore y cannot be greater than \frac{1}{4}(2n+1)^{2}, or, n^{2}+n+\frac{1}{4}; and since y is integral its greatest value must be n^{2}+n; in which case x=n+1, or n; thus, the two parts are n and n+1.

Sometimes we may use the following method:

Find the minimum value of \frac{(a+x)(b+x)}{c+x}.

Solution:

Put c+x=y; then the expression =\frac{(a-c+y)(b-c+y)}{y}=\frac{(a-c)(b-c)}{y}+y+a-c+b-c

which in turn equals

(\frac{\sqrt{(a-c)(b-c)}}{\sqrt{y}}-\sqrt{y})^{2}+a-c+b-c+2\sqrt{(a-c)(b-c)}.

Hence, the expression is a a minimum when the square term is zero; that is when y=\sqrt{(a-c)(b-c)}.

Thus, the minimum value is a-c+b-c+2\sqrt{(a-c)(b-c)}, and the corresponding value of x is \sqrt{(a-c)(b-c)}-c.

Problems for Practice:

  1. Find the greatest value of x in order that 7x^{2}+11 may be greater than x^{3}+17x.
  2. Find the minimum value of x^{2}-12x+40, and the maximum value of 24x-8-9x^{2}.
  3. Show that (n!)^{2}>n^{n} and 2.4.6.\ldots 2n<(n+1)^{n}.
  4. Find the maximum value of (7-x)^{4}(2+x)^{5} when x lies between 7 and -2.
  5. Find the minimum value of \frac{(5+x)(2+x)}{1+x}.

More later,

Nalin Pithwa.

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