2 beautiful inequalities with beautiful solutions :)

Posted on November 29, 2007 by aimingbeyond

In the book ”Old and New Inequalities” by Titu Andreescu, Gabriel Dospinescu, Vasile Cîrtoaje and Mircea Lascu the following inequality appears as problem 19, b)

[Marian Tetiva] Let $x,y,z$ be positive real numbers satisfying the condition $x^2+y^2+z^2+2xyz=1$

Prove that $x+y+z\le\displaystyle\frac32$

Solution 1:

Clearly $x,y,z\le1$ . The given relation rewrites as $(x+yz)^2=(1-y^2)(1-z^2)$ , or $x=-yz+\sqrt{(1-y^2)(1-z^2)}\le-yz+\displaystyle\frac{1-y^2+1-z^2}2=1-\frac{(y+z)^2}2$ .

Let $s=y+z$ . Then for the inequality $x+y+z\le\displaystyle\frac32$ to be true, it’s enough to have $1-\displaystyle\frac{s^2}2+s\le\frac32$ . But the last one is $-(s-1)^2\le0$ , clearly true $\blacksquare$

Solution 2:

Recall the following identity for an acute triangle: $\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=1$ . Hence we can substitute $x=\cos A$ and similar for $y,z$ . Now, $\cos$ is a concave function on $\displaystyle\left[0;\frac{\pi}2\right]$ , so, using Jensen’s Inequality we have $x+y+z=\cos A+\cos B+\cos C\le3\cos\displaystyle\frac\pi3=\frac32\blacksquare$

A beautiful solution similar in manner to the Solution 1 above, can be given for Problem 26 d) of the same book:

[Marian Tetiva] Consider positive real numbers $x,y,z$ so that $x^2+y^2+z^2=xyz$ . Prove that $xy+yz+zx\ge2(x+y+z)+9$ .

Solution 1:

By applying the AM-GM inequality, we easily deduce $\displaystyle\frac{x+y+z}3\ge\sqrt[3]{xyz}\ge3$ , hence $x+y+z\ge9$ .

We will prove the stronger inequality $xy+yz+zx\ge3(x+y+z)$ .

Write $x^2+y^2+z^2$ as $-x^2+(x^2+y^2)+(x^2+z^2)$ and deduce $xyz=x^2+y^2+z^2=-x^2+(x^2+y^2)+(x^2+z^2)\ge-x^2+2xy+2xz$ , hence $yz\ge -x+2y+2z$ . In a similar way $xz\ge -y+2x+2z$ and $xy\ge -z+2x+2y$ . Summing up these 3 inequalities we get $xy+yz+zx \ge 3(x+y+z) \blacksquare$ . Isn’t it beautiful?

None of these solutions appears in the book Here’s the solution from the book for the second problem:

Solution 2:

Since $x^2<xyz$ we get $x<yz$ and the likes. Then $xy<yz\cdot xz$ , so $1<z$ and the likes. Set $a=x-1$ , $b=y-1$ and $c=z-1$ . Clearly $a,b,c>0$ . The given condition then becomes, after expanding, $a^2+b^2+c^2+a+b+c+2=abc+ab+bc+ca$ . Set $q=ab+bc+ca$ . It is well-known that $q\le a^2+b^2+c^2$ and $\sqrt{3q}\le a+b+c$ and $abc\le\displaystyle\left(\frac q3\right)^{\frac32}=\frac{(3q)^{\frac32}}{27}$ . Then $q+\sqrt{3q}+2\le a^2+b^2+c^2+a+b+c+2=abc+ba+bc+ca\le\frac{(3q)^{\frac32}}{27}+q$ . Setting $t=\sqrt{3q}$ , we get $t+2\le\frac{t^3}{27}$ , or $(t+3)^2(t-6)\ge0$ . So $\sqrt{3q}\ge6$ , or $ab+bc+ca\ge12$ . Substituting back $x=a+1$ ,…, and expanding we get $xy+yz+zx\ge 2(x+y+z)+9\blacksquare$