MHB Is the Inequality $(a+b+c)^2+(x+y+z)^2≥1$ Always True?

[anemone]

Let $x,\,y,\,z>0$ and $a,\,b,\,c$ be real numbers such that $x^2+a^2=y^2+b^2=z^2+c^2=1$.

Prove that $(a+b+c)^2+(x+y+z)^2\ge 1$.

 

[Opalg]

Let $x,\,y,\,z>0$ and $a,\,b,\,c$ be real numbers such that $x^2+a^2=y^2+b^2=z^2+c^2=1$.

Prove that $(a+b+c)^2+(x+y+z)^2\ge 1$.

[sp]Here is a geometric approach to the problem.

[TIKZ]\clip (-6,-1) rectangle (12,12) ;
\draw [thin] (-6,0) -- (6,0) ;
\draw [thin] (0,-1) -- (0,10) ;
\coordinate [label=above left:{$P$}] (P) at (-4,3) ;
\coordinate [label=above right:{$Q$}] (Q) at (1.92,4.62) ;
\coordinate [label=right:{$R$}] (R) at (4.42,2.35) ;
\draw [thin,gray] (P) -- (3.17,3.5) ;
\draw [thin,gray] (Q) -- (0.21,2.675) ;
\draw [thin,gray] (R) -- (-1.04,3.81) ;
\coordinate [label=below:{$C$}] (C) at (0,0) ;
\coordinate [label=right:{$G$}] (G) at (0.78,3.33) ;
\coordinate [label=right:{$H$}] (H) at (2.34,10) ;
\coordinate [label=below:{$N$}] (N) at (1.75,2.55) ;
\draw [thin,gray] (H) -- (N) ;
\draw [red] (C) --(G) -- (H) ;
\foreach \point in {C,G,H}
\fill (\point) circle(2pt) ;
\draw (0,0) circle (5cm) ;
\draw (P) -- (Q) -- (R) -- cycle ;
[/TIKZ]​

Let $P = (a,x)$, $Q = (b,y)$ and $R = (c,z)$. The given conditions imply that $P,Q,R$ all lie on the upper half of the unit circle. Relabelling the points if necessary, we may assume that $Q$ lies on the arc between $P$ and $R$. Notice that the angle $PQR$ is obtuse.

The origin $C$ is the circumcentre of the triangle $PQR$. Let $G = \bigl(\frac13(a+b+c),\frac13(x+y+z)\bigr)$ be the centroid of the triangle, and let $H$ be the orthocentre (where the three perpendiculars from the vertices to the opposite sides meet). In particular, let $N$ be the point where the perpendicular from $Q$ meets the side $PR$. The fact that the angle at $Q$ is obtuse means that $N$ lies inside the unit circle on the segment $PR$, and $H$ lies outside the circle, on the opposite side of $Q$ from $N$. Therefore the distance $|CH|$ is greater than $1$.

The theory of the Euler line tells us that the points $C,G,H$ are collinear and that $|GH| = 2|CG|$, so that $|CH| = 3|CG|$. Since $G = \bigl(\frac13(a+b+c),\frac13(x+y+z)\bigr)$, it follows that $H = \bigl((a+b+c),(x+y+z)\bigr)$. But then the fact that $|CH| >1$ tells us that $(a+b+c)^2 + (x+y+z)^2 > 1.$[/sp]

 

[Albert1]

[sp]Here is a geometric approach to the problem.

[TIKZ]\clip (-6,-1) rectangle (12,12) ;
\draw [thin] (-6,0) -- (6,0) ;
\draw [thin] (0,-1) -- (0,10) ;
\coordinate [label=above left:{$P$}] (P) at (-4,3) ;
\coordinate [label=above right:{$Q$}] (Q) at (1.92,4.62) ;
\coordinate [label=right:{$R$}] (R) at (4.42,2.35) ;
\draw [thin,gray] (P) -- (3.17,3.5) ;
\draw [thin,gray] (Q) -- (0.21,2.675) ;
\draw [thin,gray] (R) -- (-1.04,3.81) ;
\coordinate [label=below:{$C$}] (C) at (0,0) ;
\coordinate [label=right:{$G$}] (G) at (0.78,3.33) ;
\coordinate [label=right:{$H$}] (H) at (2.34,10) ;
\coordinate [label=below:{$N$}] (N) at (1.75,2.55) ;
\draw [thin,gray] (H) -- (N) ;
\draw [red] (C) --(G) -- (H) ;
\foreach \point in {C,G,H}
\fill (\point) circle(2pt) ;
\draw (0,0) circle (5cm) ;
\draw (P) -- (Q) -- (R) -- cycle ;
[/TIKZ]​

Let $P = (a,x)$, $Q = (b,y)$ and $R = (c,z)$. The given conditions imply that $P,Q,R$ all lie on the upper half of the unit circle. Relabelling the points if necessary, we may assume that $Q$ lies on the arc between $P$ and $R$. Notice that the angle $PQR$ is obtuse.

The origin $C$ is the circumcentre of the triangle $PQR$. Let $G = \bigl(\frac13(a+b+c),\frac13(x+y+z)\bigr)$ be the centroid of the triangle, and let $H$ be the orthocentre (where the three perpendiculars from the vertices to the opposite sides meet). In particular, let $N$ be the point where the perpendicular from $Q$ meets the side $PR$. The fact that the angle at $Q$ is obtuse means that $N$ lies inside the unit circle on the segment $PR$, and $H$ lies outside the circle, on the opposite side of $Q$ from $N$. Therefore the distance $|CH|$ is greater than $1$.

The theory of the Euler line tells us that the points $C,G,H$ are collinear and that $|GH| = 2|CG|$, so that $|CH| = 3|CG|$. Since $G = \bigl(\frac13(a+b+c),\frac13(x+y+z)\bigr)$, it follows that $H = \bigl((a+b+c),(x+y+z)\bigr)$. But then the fact that $|CH| >1$ tells us that $(a+b+c)^2 + (x+y+z)^2 > 1.$[/sp]

Spoiler

but how can it be: $(a+b+c)^2+(x+y+z)^2=1\,\, ?$
for the question :$(a+b+c)^2+(x+y+z)^2\geq1 $
when will the equality happen ?

 

[vidyarth]

Spoiler

Since $x^2+a^2=1,$ therefore, $0<x\le1 $ and $|a|<1$ by the given conditions($x,y,z,a,b,c\in\mathbb{R}$). Similarly, $0<y,z\le1$ and $|b|,|c|<1$. Now, $$(x+y+z)^2+(a+b+c)^2=3+2(xy+yz+zx)+2(ab+bc+ca)\ge3+2(ab+bc+ca)$$ because of the positivity of $x,y,z$. Now, atleast one of $ab,bc$ or $ca$ must be positive and the minimum value of the sum of other two approaches $-2$. Note that when the negative products approach $-2$, the positive product approaches $1$.Therefore, $(x+y+z)^2+(a+b+c)^2\gt3+2(1)-2(2)=1$.

 

[Opalg]

Spoiler

but how can it be: $(a+b+c)^2+(x+y+z)^2=1\,\, ?$
for the question :$(a+b+c)^2+(x+y+z)^2\geq1 $
when will the equality happen ?

[sp]Equality can only happen in the limiting case where (in the diagram in my previous comment) $P$ and $Q$ are on the $x$-axis at the points $\pm1$. But the question stipulates that $x$, $y$ and $z$ are strictly positive. That implies that the limiting case cannot occur, and so the conclusion can be strengthened to $(a+b+c)^2+(x+y+z)^2 > 1 $.[/sp]