Giả thiết của dề bài chưa đúng, mình sửa lại thành \(cosA+cosB+cosC=\sqrt{cosA.cosB}+\sqrt{cosB.cosC}+\sqrt{cosC.cosA}\)

Đặt \(a=\sqrt{cosA},b=\sqrt{cosB},c=\sqrt{cosC}\)

Suy từ giả thiết : 

\(2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ca\right)\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}a=b=c\\a,b,c>0\end{cases}}\)

Vậy ta có \(\sqrt{cosA}=\sqrt{cosB}=\sqrt{cosC}\Rightarrow\hept{\begin{cases}cosA=cosB=cosC\\\widehat{A}+\widehat{B}+\widehat{C}=180^o\end{cases}}\)

\(\Rightarrow\widehat{A}=\widehat{B}=\widehat{C}=60^o\)

\(\Rightarrow\Delta ABC\) là tam giác đều.

Ta có : \(cos2A+2\sqrt{2}\left(cosB+cosC\right)=3\)

\(\Leftrightarrow1-2sin^2A+2\sqrt{2}.2.cos\left(\dfrac{B+C}{2}\right).cos\left(\dfrac{B-C}{2}\right)=3\)

\(\Leftrightarrow2sin^2A-4\sqrt{2}.sin\dfrac{A}{2}.cos\left(\dfrac{B-C}{2}\right)+2=0\)

\(\Leftrightarrow sin^2A-2\sqrt{2}.sin\dfrac{A}{2}.cos\left(\dfrac{B-C}{2}\right)+1=0\) 

\(\Delta\) ABC không tù nên \(cos\dfrac{A}{2}\ge cos45^o=\dfrac{\sqrt{2}}{2}\) 

Suy ra : VT \(\ge sin^2A-4.cos\dfrac{A}{2}.sin\dfrac{A}{2}.cos\left(\dfrac{B-C}{2}\right)+1=K\)

Thấy : \(K=sin^2A-2.sinA.cos\left(\dfrac{B-C}{2}\right)+cos\left(\dfrac{B-C}{2}\right)^2+1-cos\left(\dfrac{B-C}{2}\right)^2\)

\(=\left(sinA-cos\left(\dfrac{B-C}{2}\right)\right)^2+sin^2\left(\dfrac{B-C}{2}\right)\ge0\) 

Suy ra : \(VT\ge K\ge0=VP\)

 Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}sinA=cos\left(\dfrac{B-C}{2}\right)\\sin\left(\dfrac{B-C}{2}\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}sinA=cos0^o=1\\B=C\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}A=\dfrac{\pi}{2}\\B=C=\dfrac{\pi}{4}\end{matrix}\right.\)  ( do \(A+B+C=\pi\) ) 

Vậy ... 

<=> 2.cos²A - 1  + 2\(\sqrt{2}\). (cosB + cosC) = 3

<=> 2.cos²A +  2\(\sqrt{2}\). 2. cos\(\frac{B+C}{2}\). cos\(\frac{B-C}{2}\)  - 4 = 0

<=> 2. cos²A +  4\(\sqrt{2}\).sin \(\frac{A}{2}\). cos\(\frac{B-C}{2}\)  - 4 = 0 (Do  cos\(\frac{B+C}{2}\)=  cos\(\frac{180^o-A}{2}\)= sin \(\frac{A}{2}\))

Nhận xét: tam giác ABC tù nên cosA > 0;  Mà cosA \(\le\) 1   => cos²A \(\le\) cosA

Có: cos\(\frac{B-C}{2}\) \(\le\) 1

=>0 =  2. cos²A +  4\(\sqrt{2}\).sin \(\frac{A}{2}\). cos\(\frac{B-C}{2}\)  - 4 \(\le\) 2cosA +   4\(\sqrt{2}\).sin \(\frac{A}{2}\). cos\(\frac{B-C}{2}\)  - 4

= 2.(1 - 2sin² \(\frac{A}{2}\)) +  4\(\sqrt{2}\).sin \(\frac{A}{2}\)  - 4 = -2. (2sin² \(\frac{A}{2}\)-  2\(\sqrt{2}\).sin \(\frac{A}{2}\) + 1) =  -2. \(\left(\sqrt{2}sin\frac{A}{2}-1\right)^2\)\(\le\)0

=>   \(\sqrt{2}sin\frac{A}{2}-1=0\) <=> \(sin\frac{A}{2}=\frac{1}{\sqrt{2}}\)<=> A/2 = 45^o

=> góc A = 90^o

Dấu "=" xảy ra  <=> cos\(\frac{B-C}{2}\) = 1 => B - C = 0 => B = C mà A = 90^o

=> B = C = 45^o

vậy..........

\(\Rightarrow \tan A+\tan C=2\tan B\)

\(\Leftrightarrow \frac{\sin\left ( A+C \right )}{\cos A\cos C}=2\cdot\frac{\sin\left ( A+C \right )}{\cos B}\\\)

\(\Rightarrow \cos B=2\cos A\cos C\)

\(\Leftrightarrow 2\cos B=\cos(A-C)\)

\(\left (\cos A+\cos C \right )^2=\cos^2 A+\cos^2 C+2\cos A\cos C\\=\frac{\cos2A+\cos2C}{2}+1+\cos B\\=-\cos(B)\cos(A-C)+1+\cos B \\=-2\cos^2B+\cos B+1 \le \frac{9}{8}\\\Rightarrow \cos A+\cos C\le \frac{3\sqrt2}{4}\)

Chứng minh hoàn tất.

use mot cay gay

Ta chứng minh chiều nghịch:

Khi tam giác ABC đều, góc A=gócB=gócC=60*

Khi đó cosA+cosB+cosC=3/2(đpcm)

Ta chứng minh chiều thuận

Ta chứng minh cosA+cosB+cosC≤3/2

Thật vậy:

 Mà theo gt, cosA+cosB+cosC=3/2

nên ta có tam giác ABC đều(đpcm)

vẽ AD,BE, CF là các đường cao của tam giác ABC

\(\cos A=\sqrt{\cos BAE\cdot\cos CAF}=\sqrt{\frac{AE}{AB}\cdot\frac{AE}{AC}}=\sqrt{\frac{AF}{AB}\cdot\frac{AE}{AC}}\le\frac{1}{2}\left(\frac{AF}{AB}+\frac{AE}{AC}\right)\)

ta có \(\cos A\le\frac{1}{2}\left(\frac{AF}{AB}+\frac{AE}{AC}\right)\left(1\right)\)

tương tự \(\cos B\le\frac{1}{2}\left(\frac{BF}{AB}+\frac{BD}{BC}\right)\left(2\right);\cos C\le\frac{1}{2}\left(\frac{CD}{BC}+\frac{CE}{AC}\right)\left(3\right)\)

do đó \(\cos A+\cos B+\cos C\le\frac{1}{2}\left(\frac{AF}{AB}+\frac{AE}{AC}+\frac{BF}{AB}+\frac{BD}{BC}+\frac{CD}{BC}+\frac{CE}{AC}\right)\)

\(\Rightarrow\cos A+\cos B+\cos C\le\frac{1}{2}\left(\frac{AF}{AB}+\frac{BF}{AB}+\frac{AE}{AC}+\frac{CE}{AC}+\frac{BD}{BC}+\frac{CD}{BC}\right)\)

\(\Rightarrow\cos A+\cos B+\cos C\le\frac{3}{2}\)

dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\frac{AF}{AB}=\frac{AE}{AC}\\\frac{BF}{AB}=\frac{BD}{BC}\\\frac{CD}{BC}=\frac{CE}{AC}\end{cases}}\Leftrightarrow AB=AC=BC\)

do vậy cosA+cosB+cosC=3/₂ <=> AB=AC=BC <=> tam giác ABC đều

Cách khác khỏi phải dùng hình học :v

\(A=\cos A+\cos B+\cos C\)

\(=\left(\cos A+\cos B\right)\cdot1+\sin A\cdot\sin B-\cos A\cdot\cos B\)

\(\le\frac{1}{2}\left[\left(\cos A+\cos B\right)^2+1\right]+\frac{1}{2}\left(\sin^2A+\sin^2B\right)-\cos A\cdot\cos B\)

\(=\frac{1}{2}\left(\cos^2A+\sin^2A+\cos^2B+\sin^2B\right)+\frac{1}{2}\)

\(=\frac{3}{2}\)

ez Problem :v