Spring 2018, problem 59
Comments
Let us call $F(a,b,c)$ the function to be minimized. It is a homogeneous function of degree 0 say, for any $t$, it is $F(ta,tb,tc)=F(a,b,c)$. Thus we can confine ourselves to work with $(x,y,z)$ in the triangle $T:=\{x+y+z=1;\ x,y,z\ge0\}$; and we will see that the minimum is when the variables all take the value $\frac13$.
On the boundary of $T$ the function $F$ is bigger than 1: e.g. for $a=0$ the function becomes $1+2\frac{(b-c)^2}{(b^2+c^2)}\ge1$. Thus we can apply the Lagrange method, searching for minima at the interior of the triangle. Better, we will search for interior maxima on $T$ of the function $6-F(a,b,c)$ that, as we will show, can be written on the form $g(a)+g(b)+g(c)$, the function $g$ being defined by $g(x):=\frac1{(1-x)^2+x^2}$.
In fact $2-\frac{(b+c−a)^2}{(b+c)^2+a^2}=\frac{(a+b+c)^2}{(b+c)^2+a^2}$ that on $a+b+c=1$ becomes $\frac1{(1-a)^2+a^2}$, say $g(a)$; the other terms can obviously be treaten in the same way.
Now the Lagrange method gives for the estremal points $(a_0,b_0,c_0)$ the equation $g'(a_0)=g'(b_0)=g'(c_0)$ that, joint with $a_0+b_0+c_0=1$ implies $a_0=b_0=c_0=\frac13$.
Added on february 19
May be the last point deserves more details.
For any real $k$ the equation $g'(x)=k$ has at most two solutions; in particular if $g'(a)=g'(b)=g'(c)$ then at least two among $a,b,c$ must coincide; let e.g. $a=b$. Being on the triangle $T$ we must have $c=1-2a$, and obviously $a\in(0,\frac12)$. Our goal is $a=\frac13$ (thus also $c=a=\frac13$). And in fact, rewriting $g'(a)=g'(c)$ we get $g'(a)-g'(1-2a)=0$.
This imlpies $a=c$ because on the interval $(0,\frac12)$ the function $h(x):=g'(x)-g'(1-2x)$ starts from $h(0)=4$, then increases on a small interval, then decreases; thus it vanishes only on $x_0:=\frac13$.
@Phiboyd2
The idea that a symetric function ( that stays unchanged when exchanging the variables) should have a maximum or minimum for equal values of the variables is very attractive but generally wrong for polynomials of degree 4 or more as proved by Buniakovsky in 1854.It is true for degrees less than 4. (here the degree of the polynomial is 6 so one cannot use this argument) a nice example is the following: P(x,y) = {x^2 +(1-y)^2} {y^2 +(1-x)^2} P is alwways positive or null and is equal to zero for (1,0) or (0,1).
@Claudio
Your remark that the problem can be confined to the case a+b+c= 1 is indeed crucial. Then you can get an immediate proof of the inequality using Jensen's theorem and the convexity of the function:
f(x) = (1-2x)^2 / {(1-x)^2 + x^2}
Thanks, Mike. You are right, I did't think to the Jensen's inequality, thus my solution is unnecessarily complicayted; but it was a first attempt... Can I suggest you of writing down the simpler solution?
With the assumption a+b+c = 1 we have to prove that f(a) +f(b)+f(c) larger or equal to 3/5 with f(x) defined as above. f(x) is convex on the segment [ (3-sqt(3))/6 , (3+sqrt(3))/6] because its second derivative is non negative.
So if a,b and c are in this segment we can use Jensen's theorem which gives: f(a)+f(b)+f(c) larger or equal than 3f( (a+b+c)/3)=3f(1/3)=3/5.
Now suppose WLOG that a is larger than (3+sdrt(3))/6 then b+c is smaller than 1- (3+sqrt(3))/6 which is precisely the other bound (3- sqrt(3))/6; We therfore conclude that if a is outside the convex region, then b and c are also outside the convex region.But f(x) is larger than 1/2 outside this segment ,hence f(a) +f(b)+f(c) is larger than 3/2 in that case
In conclusion whatever the values of a,b,c in the segment (0,1) with hte condition a+b+c=1, we have always 3/5 as the lower bound.
@Phiboyd2
The idea that a symetric function ( that stays unchanged when exchanging the variables) should have a maximum or minimum for equal values of the variables is very attractive but generally wrong for polynomials of degree 4 or more as proved by Buniakovsky in 1854.It is true for degrees less than 4. (here the degree of the polynomial is 6 so one cannot use this argument) a nice example is the following: P(x,y) = {x^2 +(1-y)^2} {y^2 +(1-x)^2} P is alwways positive or null and is equal to zero for (1,0) or (0,1).
@Claudio
Your remark that the problem can be confined to the case a+b+c= 1 is indeed crucial. Then you can get an immediate proof of the inequality using Jensen's theorem and the convexity of the function:
f(x) = (1-2x)^2 / {(1-x)^2 + x^2}
- Call the function on the LHS to be f(a,b,c).
- Since all variables are independent, fix b and c.Let the variable 'a' vary alone.So if we want to find minimum of f(a,b,c) with fixed b and c we differentiate f partially with respect to a and equate it with 0 to get a=β(say)..
3.Now,let the variable b vary alone keeping others fixed and again differentiating f partially with respect b and equatng it with 0 in order to find minimum of f(a,b,c).By symmetry b must be equals to β.In fact the last variable c must be equals to β(while doing same process with c).
4.These will tell us that the all variables will have to be equal(for minimum value of f(a,b,c) )
because if we intersect the conditions for minimum value of f(a,b,c) with respect to individual independent variables we will get the condition which will give global minimum of f(a,b,c). 5.Thus putting a=b=c we will get 3/5 as a minimum of f(a,b,c).
My solution in 3 simple steps
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Since the left side is a homogenous function F(a,b,c)=F(ta,tb,tc) we can restrict a,b and c to be smaller or equal to 1 and recognize that F(1,1,1) = 3/5
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Now we set * a+b+c = 3 * and substitute/replace yielding the expression (3-2c)^2/ ( (3-c)^2+ c^2)) + (3-2a)^2/( ( 3-a)^2 + a^2)) + (3-2b)^2/ ( (3-b)^2 + b^2)). These are rational functions. I WILL SHOW THAT EACH OF THE THREE TERMS IS always LARGER THAN 1/5.
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Expanding the quadratic functions and doing the PARTIAL FRACTION DIVISION yields : (9 - 12c + 4 c^2 ) / ( 9 - 6c+2c^2) = 2 - 9 / ( 2 c^2 - 6c + 9 ) , the same for a and b. The quadratic polynomial in the denominator takes its minimum value at c=1 , the whole function is therefore a strictly decreasing function and falls from 1 at c=0 to its minimum value of 1/5 at c=0. Adding the three terms 3x1/5=3/5 completes the proof.
Let us assume for any $a,b,c \in \mathbb{R^{+}}$ we have $a \le b \le c$. Now, let $b = x \cdot a, c = y \cdot a$ such that $1 \le x \le y$ holds for $x,y \in \mathbb{R^{+}}$. This now allows us to transform the above 3-variable expression into a 2-variable version:
$\frac{(b+c-a)^2}{(b+c)^2 + a^2} + \frac{(c+a-b)^2}{(c+a)^2 + b^2} + \frac{(a+b-c)^2}{(a+b)^2 + c^2} = \frac{(x+y-1)^2a^2}{[(x+y)^2 + 1]a^2} + \frac{(y+1-x)^2 a^2}{[(y+1)^2 + x^2]a^2} + \frac{(1+x-y)^2 a^2}{[(1+x)^2 + y^2]a^2}$
or $f(x,y) = [1 - \frac{2(x+y)}{(x+y)^2 + 1}] + [1 - \frac{2x(y+1)}{(y+1)^2 + x^2}] + [1 - \frac{2y(x+1)}{(x+1)^2 + y^2}].$
If $1 \le x \le y$ holds for $x,y \in \mathbb{R^{+}}$, then for some fixed $x=x_{0}$ we have $lim_{y \rightarrow \infty} f(x_{0},y) = 3$ (which is the maximum value for $f$). Also, there exists only one critical point in the first-quadrant of the $xy$-plane: $(x,y) = (1,1)$. Evaluating $f$ at this critical point gives:
$f(1,1) = 3 \cdot (1 - \frac{4}{5}) = 3 \cdot \frac{1}{5} =\boxed{ \frac{3}{5}}$.
Since $\frac{3}{5}$ < $3$, this is our minimum value.