Kẽ Bx // AC cắt AM tại Q.

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{MA}{AQ}=\dfrac{MC}{BC}\\\dfrac{MC}{MB}=\dfrac{AC}{BQ}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}MA.BC=MC.AQ\\MC.BQ=MB.AC\end{matrix}\right.\)

\(\Rightarrow MB.AC+MC.AB=MC.BQ+MC.AB=MC\left(AB+BQ\right)>MC.AQ=MA.BC\)

sp làm đúng lúc con cần luôn :V Tks Sp

Dễ dàng chứng minh IC,IA,IB lần lượt vuông góc với DE,EF,DF

nên \(DE=2DS=2CD.\sin\dfrac{C}{2}=\left(a+b-c\right).\sin\dfrac{C}{2}\)

tương tự với EF và DF,ta cần chứng minh :

\(\sum\dfrac{\left(a+b-c\right).\sin\dfrac{C}{2}}{\sqrt{ab}}\le\dfrac{3}{2}\)

có bổ đề :\(\sin\dfrac{A}{2}\le\dfrac{a}{b+c}\) ( H.b)( tự chứng minh)

nên BĐT cần chứng minh : \(\sum\dfrac{\left(a+b-c\right).c}{\left(a+b\right)\sqrt{ab}}\le\dfrac{3}{2}\)

AM-GM: \(\left(a+b\right)\sqrt{ab}\ge2\sqrt{ab}.\sqrt{ab}=2ab\)

Tương tự: \(VT\le\sum\dfrac{\left(a+b-c\right)c}{2ab}=\dfrac{\sum ab\left(a+b\right)-\sum a^3}{2abc}\)

Áp dụng BĐT schur: \(ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)\le a^3+b^3+c^3+3abc\)

( cm : \(\Leftrightarrow\sum a\left(a-b\right)\left(a-c\right)\ge0\) và ta có thể giả sử \(a\ge b\ge c\)...Google để chi tiết )

\(\Rightarrow VT\le\dfrac{3abc}{2abc}=\dfrac{3}{2}\)

Dấu = xảy ra khi a=b=c.( a,b,c>0)

P/s: để ý rằng \(\sum\dfrac{\left(a+b-c\right)c^2}{2abc}=\sum\dfrac{\left(b^2+c^2-a^2\right)a}{2abc}=\sum\dfrac{b^2+c^2-a^2}{2bc}=\sum\cos A\)

Điều kiện: \(\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\y\ge\dfrac{3}{4}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}2x\sqrt{y}+y\sqrt{x}=3\sqrt{4y-3}\left(1\right)\\2y\sqrt{x}+x\sqrt{y}=3\sqrt{4x-3}\left(2\right)\end{matrix}\right.\)

Lấy (1) - (2) ta được

\(x\sqrt{y}-y\sqrt{x}+3\left(\sqrt{4x-3}-\sqrt{4y-3}\right)=0\)

\(\Leftrightarrow\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)+\dfrac{12\left(x-y\right)}{\sqrt{4x-3}+\sqrt{4y-3}}=0\)

\(\Leftrightarrow\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{xy}+\dfrac{12\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{4x-3}+\sqrt{4y-3}}\right)=0\)

\(\Leftrightarrow x=y\)

\(\Rightarrow x\sqrt{x}=\sqrt{4x-3}\)

\(\Leftrightarrow x^3=4x-3\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+x-3\right)=0\)

Tới đây bí

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\y\ge\dfrac{3}{4}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}2x\sqrt{y}+y\sqrt{x}=3\sqrt{4y-3}\left(1\right)\\2y\sqrt{x}+x\sqrt{y}=3\sqrt{4x-3}\left(2\right)\end{matrix}\right.\)

Lấy \(\left(1\right)-\left(2\right)\) theo từng vế:

\(2x\sqrt{y}+y\sqrt{x}-2y\sqrt{x}-x\sqrt{y}=3\sqrt{4y-3}-3\sqrt{4x-3}\)

\(x\sqrt{y}-y\sqrt{x}-3\left(\sqrt{4y-3}-\sqrt{4x-3}\right)=0\)

\(\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)+\dfrac{12\left(x-y\right)}{\sqrt{4x-3}+\sqrt{4y-3}}=0\)

\(\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{xy}+\dfrac{12\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{4x-3}-\sqrt{4y-3}}\right)=0\)

\(\Leftrightarrow x=y\)

\(\Rightarrow x\sqrt{x}=\sqrt{4x-3}\)

\(\Rightarrow x^3=4x-3\)

\(\left(x-1\right)\left(x^2+x+3\right)=0\)

Dễ thấy: \(x^2+x+3=x^2+2x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{11}{4}\)

\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}>0\forall x\)

\(\Rightarrow x=1\)

\(\Rightarrow y=1\)

\(\dfrac{1}{\cos^2x}=1+\tan^2x=1+4=5\)\(\Rightarrow\cos^2x=\dfrac{1}{5}\Rightarrow\cos x=\dfrac{\sqrt{5}}{5}\)\(\Rightarrow\sin x=\tan x.\cos x=\left(-2\right).\dfrac{\sqrt{5}}{5}=\dfrac{-2\sqrt{5}}{5}\)

\(A=\dfrac{\dfrac{-4\sqrt{5}}{5}+\dfrac{\sqrt{5}}{5}}{\dfrac{\sqrt{5}}{5}+\dfrac{6\sqrt{5}}{5}}\)\(=\dfrac{-3}{7}\)

Hint: \(M=3-Σ\dfrac{a^2+a+1}{a^2+2a+1}\)

and \(a=b=c=\sqrt 3-1; Max_{M}=\sqrt3 -1\) :))

Mấy bạn giải giúp mik nha

Guể :v t nhớ làm bài này rồi mà :v

Đặt \(x=\dfrac{bc}{a^2};y=\dfrac{ac}{b^2};z=\dfrac{ab}{c^2}\)\(\Rightarrow\left\{{}\begin{matrix}abc=1\\a,b,c>0\end{matrix}\right.\)

Và \(BDT\Leftrightarrow\dfrac{a^4}{b^2c^2+a^2bc+a^4}+\dfrac{b^4}{a^2c^2+ab^2c+b^4}+\dfrac{c^4}{a^2b^2+abc^2+c^4}\ge1\)

Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:

\(VT\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{b^2c^2+a^2bc+a^2c^2+ab^2c+a^2b^2+abc^2+a^4+b^4+c^4}\)

Cần chứng minh \(\dfrac{\left(a^2+b^2+c^2\right)^2}{b^2c^2+a^2bc+a^2c^2+ab^2c+a^2b^2+abc^2+a^4+b^4+c^4}\ge1\)

\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2\ge b^2c^2+a^2bc+a^2c^2+ab^2c+a^2b^2+abc^2+a^4+b^4+c^4\)

\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)\ge b^2c^2+a^2bc+a^2c^2+ab^2c+a^2b^2+abc^2+a^4+b^4+c^4\)

\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2\ge ab^2c+a^2bc+abc^2\)

\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2\ge abc\left(a+b+c\right)\) *Đúng theo AM-GM*

uh bài này làm rồi, tại lúc đó đầu hơi ngu nên không nhớ ra, thông cảm nhé

Đặt \(\left\{{}\begin{matrix}S_{BDF}=x\\S_{CEF}=y\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}S_{ADF}=3x\\S_{AEF}=4y\\S_{ABF}=4x\\S_{ACF}=5y\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}S_{ABE}=4x+4y=\dfrac{4}{5}S_{ABC}\\S_{ACD}=3x+5y=\dfrac{3}{4}S_{ABC}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}20x+20y=4S_{ABC}\\12x+20y=3S_{ABC}\end{matrix}\right.\)

\(\Rightarrow S_{ABC}=8x=2S_{ABF}\)

Xong

Giải ra cả nùi số đằng sau dấu phẩy luôn.

\(M=\sqrt{\dfrac{a}{b+c+2a}}+\sqrt{\dfrac{b}{c+a+2b}}+\sqrt{\dfrac{c}{a+b+2c}}\)

\(\le\dfrac{1}{4}+\dfrac{a}{b+c+2a}+\dfrac{1}{4}+\dfrac{b}{c+a+2b}+\dfrac{1}{4}+\dfrac{c}{a+b+2c}\)

\(\le\dfrac{3}{4}+\dfrac{1}{4}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{b}{b+c}+\dfrac{b}{a+b}+\dfrac{c}{c+a}+\dfrac{c}{b+c}\right)\)

\(=\dfrac{3}{4}+\dfrac{1}{4}.\left(1+1+1\right)=\dfrac{3}{2}\)

Hung nguyen cho hỏi dòng 2 sao ra được v

\(S=\sum\limits^{121}_2\left(\dfrac{1}{x\sqrt{\left(x-1\right)}+\left(x-1\right)\sqrt{x}}\right)\)

\(S=0,9090909091\)

Sao mình thấy nó kì kì