Spring 2018, problem 59

Prove that

$ \frac{\left(b+c-a\right)^{2}}{\left(b+c\right)^{2}+a^{2}}+\frac{\left(c+a-b\right)^{2}}{\left(c+a\right)^{2}+b^{2}}+\frac{\left(a+b-c\right)^{2}}{\left(a+b\right)^{2}+c^{2}}\geq\frac35$

for any positive real numbers $ a$, $ b$, $ c$.

Comments

2 months ago

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.

9 months ago

@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?

Claudio 9 months ago

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.

michel 9 months ago
9 months ago

@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}

9 months ago

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 $x_0:=\frac13$.

8 months ago

My solution in 3 simple steps

  1. 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

  2. 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.

  3. 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.

3 months ago
  1. Call the function on the LHS to be f(a,b,c).
  2. 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).
9 months ago

This seems like a bit of deja vu !

The following comment extracted from the previous problem 58 states :

                        The function M(a,b) is minimized when both of its arguments equal each other.        (i)

If we apply this to this problem replacing M(a,b) by f(a,b,c) then with a=b=c the expression takes the value 3/5 - solved.

However, as I previously mentioned the statement (i) does need some justification.

I am beginning to wonder if as a general rule a function of n variables where the value of the function is the same no matter how the variables are positioned implies that the minimum, (or maximum as the case may be) occurs when all the variables have the same value.

e.g if f(x1, x2, x3, x4) = f(x1, x2, x4, x3) = f(x1, x3, x2, x4) .......... then the minimum occurs when x1=x2=x3=x4

The property you quote as (1) does not refer to a general function $M$: it just apply when the function $M$ is defined as the maximum of two functions, $f,g$. Furthermore, as I stated twice in the comments to Problem 58 (and one of the explanations was dedicated to you), the property $f(P)=g(P)$ holds true unless the minimum point $P$ is a local minimum of one among $f,g$ (no matter the number of variables). If you need further details, let me know.

Claudio 9 months ago

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