In Romanian Journal of Mathematics "Gazeta Matematica - Seria B", no 2/2022, on page 60, there is an article on Problem 28 003, which appeared in GMB 2/2021 (page 107) and was solved in GMB 9/2021 (page 404). Here we demonstrate LEMA 4, which the author states "results by direct calculation"
In translation, for which we thank DeepL, reads as follows;
" For any positive numbers $a$, $b$, $c$ there is equivalence
$$a^2+b^2+c^2+abc=4\;\;\Leftrightarrow\;\;\frac{a}{2a+bc}+\frac{b}{2b+ca}+\frac{c}{2c+ab}=1." \tag{E}$$
Solution CiP
To obtain homogeneous algebraic expressions, we use the notation $abc :=d^2$. Then the condition on the left-hand side of the (E) is expressed
$$a^2+b^2+c^2+d^2=4. \tag{1}$$
Further we have $ab=\frac{d^2}{c}$ and the analogues, so the expression on the right-hand side of (E) is
$$R:=\frac{a}{2a+bc}+\dots \;=\frac{a}{2a+\frac{d^2}{a}}+\frac{b}{2b+\frac{d^2}{b}}+\frac{c}{2c+\frac{d^2}{c}}$$
so
$$R=\frac{a^2}{d^2+2a^2}+\frac{b^2}{d^2+2b^2}+\frac{c^2}{d^2+2c^2} \tag{2}.$$
Both (1) and (2) are now homogeneous expressions. (1) is symmetric, while (2) is symmetric only with respect to $\{a,b,c\}$, so we will consider $d$ as a parameter.
Let's calculate the expression $R$ by bringing it in (2) to the same common denominator. The calculation will push itself forward. We'll substitute everywhere possible $abc \rightarrow d;\;a^2b^2c^2 \rightarrow \;d^2$.
The denominator of $R$ is
$$Q:=(2a^2+d^2)(2b^2+d^2)(2c^2+d^2)=$$
$=(4a^2b^2+2a^2d^2+2b^2d^2+d^4)(2c^2+d^2)=8a^2b^2c^2+4a^2c^2d^2+\dots +2b^2d^4+d^6.$
Finally we get
$$Q=d^2[d^4+2(a^2+b^2+c^2+4)d^2+4(a^2b^2+b^2c^2+c^2a^2)].$$
The numerator of $R$ is, more thickly
$$P:=a^2(2b^2+d^2)(2c^2+d^2)+b^2(2c^2+d^2)(2a^2+d^2)+c^2(2a^2+d^2)(2b^2+d^2)$$
but in the end we find
$$Q=d^2[(a^2+b^2+c^2+12)d^2+4(a^2b^2+b^2c^2+c^2a^2)].$$
In conclusion
$$R=\frac{P}{Q}=\frac{(a^2+b^2+c^2+12)d^2+4(a^2b^2+b^2c^2+c^2a^2)}{d^4+2(a^2+b^2+c^2+4)d^2+4(a^2b^2+b^2c^2+c^2a^2)}.\;\tag{3}$$
From relation (3) we obtain that (where the sum $\sum$ extends to $a,b,c$)
$$R=1\;\Leftrightarrow\;(\sum a^2+12)d^2=d^4+2(\sum a^2+4)d^2$$
$$\Leftrightarrow\;\sum a^2+12=d^2+2\sum a^2+8$$
$$\Leftrightarrow\;\;\sum a^2+d^2=4$$
so we got precisely condition (1). This demonstrates the required equivalence.
$\blacksquare$
REMARK
As Mr. Leo Giugiuc stated, in a personal message, the implication $\Rightarrow$ in (E) can be proved using trigonometric substitutions.
Indeed, from $a^2+b^2+c^2+abc=4$ it result $\;a^2,b^2,c^2\; \leq 4$, so let's put
$$a=2\sin \alpha,\;b=2 \sin \beta,\;\;\alpha, \beta\;\in (0, \frac{\pi}{2}].$$
Then $c$ is the positive solution of the second degree equation
$$c^2+4c \sin \alpha \cdot \sin \beta+4\sin^2 \alpha+4\sin^2 \beta-4=0,$$
for which we have $\Delta^{'}_c=4\sin^2\alpha \cdot \sin^2 \beta -4\sin^2 \alpha-4\sin^2 \beta+4=4(1-\sin^2\alpha)(1-\sin^2 \beta)=4 \cos^2 \alpha \cdot\cos^2 \beta),$
$$c=\frac{-2 \sin \alpha \cdot \sin \beta +\sqrt{\Delta^{'}_c}}{1}=2\cos \alpha \cdot \cos \beta - 2 \sin \alpha \cdot \sin \beta=2\cos (\alpha+\beta).$$
Further we have $\frac{a}{2a+bc}=\frac{2\sin \alpha}{4\sin \alpha+2\sin \beta \cdot 2\cos(\alpha+\beta)}=\frac{\sin \alpha}{2(\sin\alpha+\sin \beta \cdot \cos(\alpha+\beta))}=$
$=\frac{\sin \alpha}{2(\sin \alpha +\cos \alpha \cdot \cos \beta \cdot \sin \beta-\sin \alpha \cdot \sin^2 \beta)}=\frac{\sin \alpha}{2(\cos \alpha \cdot \cos \beta \cdot \sin \beta+\sin \alpha(1-\sin^2 \beta))}=\frac{\sin \alpha}{2\cos \beta(\cos \alpha \cdot \sin \beta+\sin \alpha \cdot \cos \beta}.$
So, repeating the same kind of calculations, we got
$$\frac{a}{2a+bc}=\frac{\sin \alpha}{2\cos \beta \cdot \sin (\alpha+\beta)},\;\frac{b}{2b+ac}=\frac{\sin \beta}{2\cos \alpha \cdot \sin(\alpha+\beta)},\;\frac{c}{2c+ab}=\frac{\cos(\alpha+\beta)}{2\cos\alpha \cdot \cos \beta}.$$
Then
$R=\frac{\sin \alpha}{2\cos \beta \cdot \sin(\alpha+\beta)}+\frac{\sin \beta}{2 \cos \alpha \cdot \sin(\alpha+\beta)}+\frac{\cos(\alpha+\beta)}{2\cos \alpha \cdot \cos \beta}=\frac{\sin \alpha \cdot \cos \alpha+\sin \beta \cdot \cos \beta}{2\cos \alpha \cdot \cos \beta \cdot \sin(\alpha+\beta)}+\frac{\cos(\alpha+\beta)}{2\cos \alpha \cdot \cos \beta}=$
$=\frac{\sin 2\alpha+\sin 2\beta}{4\cos \alpha \cdot \cos \beta \cdot \sin(\alpha+\beta) }+\frac{\cos(\alpha+\beta)}{2\cos \alpha \cdot \cos \beta}=\frac{2\sin (\alpha+\beta)\cdot \cos(\alpha-\beta)}{4\cos \alpha \cdot\cos \beta \cdot \sin(\alpha+\beta)}+\frac{\cos(\alpha+\beta)}{2\cos \alpha \cdot \cos \beta}=\frac{\cos(\alpha-\beta)}{2\cos \alpha \cdot \cos \beta}+\frac{\cos(\alpha+\beta)}{2\cos \alpha \cdot \cos \beta}=$
$=\frac{\cos(\alpha -\beta)+\cos(\alpha+\beta)}{2\cos \alpha \cdot \cos \beta}=\frac{2\cos \alpha \cdot \cos \beta}{2\cos \alpha \cdot \cos\beta}=1$.
END REM
BIRO ISTVAN makes substitutions $x=\frac{bc}{a},\;y=\frac{ca}{b},\;z=\frac{ab}{c}$ and has a simpler calculation to make.
LEO GIUGIUC points to another inequality that starts from the same hypothesis.
Kunihiko Chikaya makes an equally difficult calculation but proves both implications.
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