[Toán 11] Tìm $\min$ của $P = \tan A + \tan B + \tan C$

T

trang_dh

[TEX]tanA+tanB + tanC=\frac{sinA}{cosA} + \frac{sinB}{cosB} +\frac{sinC}{cosC} [/TEX]

[TEX]=\frac{sinA .cosB+ sinB cosA}{cosA.cosB}+\frac{sinC}{cosC}[/TEX]

[TEX]=\frac{sin(A+B)}{cosA.cosB}+\frac{sinC}{cosC}[/TEX]

[TEX]=\frac{sin(\pi-C).cos(\pi-A-B)+ sinC.cosA.cosB}{cosA.cosB.cosC}[/TEX]

[TEX]\frac{sinC.[-cos(A+B)+cosA.cosB]}{cosA.cosB.cosC}[/TEX]

[TEX]\frac{sinC.sinA.sinB}{cosA.cosB.cosC}[/TEX]

[TEX]=tanA .tanB .tanC[/TEX]

[TEX]\Rightarrow tanA+tanB + tanC=tanA .tanB .tanC[/TEX]

lại có
[TEX]tanA+tanB + tanC\geq3 \sqrt[3]{(tanA .tanB .tanC)}=3 \sqrt[3]{tanA .tanB .tanC}[/TEX]


[TEX]\Leftrightarrow (tanA +tanB +tanC)^3\geq27[/TEX]

[TEX]\Leftrightarrow (tanA +tanB +tanC)\geq\sqrt[3]{27}[/TEX]

dau bang xay ra khi tana=tanb=tanc

vay khi tam giac abc deu
 
N

noinhobinhyen

hãy cmr

$tanA \geq 4(A-\dfrac{\pi}{3})+\sqrt{3}$

đơn giản thôi xét hàm $f(x) = tanx-4(x-\dfrac{\pi}{3}) - \sqrt{3}$

tìm $minf(x) = 0$ bằng cách xét f'(x) = 0 ý.....

vậy $tanA+tanB+tanC \geq 4(A+B+C-\pi)+3\sqrt{3} = 3\sqrt{3}$
 
T

tranvanhung7997


Do $A + B + C = \pi \leftrightarrow A + B = \pi - C \leftrightarrow tan(A + B) = tan(\pi - C)$
$\leftrightarrow \dfrac{tanA + tanB}{1 - tanA.tanB} = - tanC \leftrightarrow tanA + tanB = - tan C + tanA.tanB.tanC$
$\leftrightarrow tanA + tanB + tanC = tanA.tanB.tanC$
Do tam giác ABC nhọn => tanA , tanB , tan C > 0
Áp dụng BĐT Cô Si:
$tanA + tanB + tanC \ge 3\sqrt[3]{tanA.tanB.tanC} = 3\sqrt[3]{tanA + tanB + tanC }$
=> $(tanA + tanB + tanC )^3 \ge 27(tanA + tanB + tanC )$
=> $tanA + tanB + tanC \ge 3\sqrt[]{3}$
Dấu = có <=> tam giác ABC đều
 
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