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Bài 1:
\(\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ac}{b}\ge a+b+c\) với a,b,c > 0
Áp dụng BĐT Chauchy cho 2 số không âm, ta có:
\(\dfrac{bc}{a}+\dfrac{ac}{b}=c\left(\dfrac{b}{a}+\dfrac{a}{b}\right)\ge c\sqrt{\dfrac{b}{a}.\dfrac{a}{b}}=2c\)
\(\dfrac{ac}{b}+\dfrac{ab}{c}=a\left(\dfrac{c}{b}+\dfrac{b}{c}\right)\ge a\sqrt{\dfrac{c}{b}.\dfrac{b}{c}}=2a\)
\(\dfrac{ab}{c}+\dfrac{bc}{a}=b\left(\dfrac{a}{c}+\dfrac{c}{a}\right)\ge b\sqrt{\dfrac{a}{c}.\dfrac{c}{a}}=2b\)
Cộng vế theo vế ta được:
\(2\left(\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ac}{b}\right)\ge2\left(a+b+c\right)\)
\(\Leftrightarrow\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ac}{b}\ge a+b+c\)
Bài 1:
Áp dụng BĐt cauchy dạng phân thức:
\(\dfrac{1}{2x+y}+\dfrac{1}{x+2y}\ge\dfrac{4}{3\left(x+y\right)}\)
\(\Rightarrow\left(3x+3y\right)\left(\dfrac{1}{2x+y}+\dfrac{1}{x+2y}\right)\ge\left(3x+3y\right).\dfrac{4}{3x+3y}=4\)
dấu = xảy ra khi 2x+y=x+2y <=> x=y
Bài 2:
ta có: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\ge\dfrac{4^2}{a+b+c+d}=\dfrac{16}{a+b+c+d}\)(theo BĐt cauchy-schwarz)
\(\Rightarrow\dfrac{1}{a+b+c+d}\le\dfrac{1}{16}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\right)\)
Áp dụng BĐT trên vào bài toán ta có:
\(A=\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{16}\left(\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{1}{c}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{c}\right)\)\(A\le\dfrac{1}{16}.4\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
......
dấu = xảy ra khi a=b=c
Bài 2:
Áp dụng BĐT cauchy cho 2 số dương:
\(a^2+1\ge2a\)
\(\Leftrightarrow\dfrac{a}{a^2+1}\le\dfrac{a}{2a}=\dfrac{1}{2}\)
thiết lập tương tự:\(\dfrac{b}{b^2+1}\le\dfrac{1}{2};\dfrac{c}{c^2+1}\le\dfrac{1}{2}\)
cả 2 vế các BĐT đều dương ,cộng vế với vế,ta có dpcm
dấu = xảy ra khi a=b=c=1
5)
a)
Có 3x+y = 1
\(\Rightarrow x+x+x+y=1\)
Áp dụng bất đẳng thức bunhiacopxki ta có :
\(\left(x^2+x^2+x^2+y^2\right)\left(1^2+1^2+1^2+1^2\right)\ge\left(x+x+x+y\right)^2\)
\(\Rightarrow3x^2+y^{2^{ }}.4\ge\left(3x+y\right)^2\)
\(\Rightarrow3x^2+y^2\ge\dfrac{1}{4}\)
b)
Áp dụng bất đẳng thức AM - GM ta có :
\(\left[{}\begin{matrix}a^2+1^2\ge2a\\b^2+1^2\ge2b\\c^2+1^2\ge2c\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left(a+1\right)^2\ge4a^{ }\\\left(b+1\right)^2\ge4b^{ }\\\left(c+1\right)^2\ge4c^{ }\end{matrix}\right.\)
\(\Rightarrow\left(a+1\right)^2\left(b+1\right)^2\left(c+1\right)^2\ge4a^{ }.4b.4c^{ }\)
\(\Rightarrow\left(a+1\right)^2\left(b+1\right)^2\left(c+1\right)^2\ge64a^{ }bc^{ }\)
\(\Rightarrow\left(a+1\right)^2\left(b+1\right)^2\left(c+1\right)^2\ge64abc\)
\(\Rightarrow\left(a+1\right)^2\left(b+1\right)^2\left(c+1\right)^2\ge64\)
\(\Rightarrow\left(a+1\right)^{ }\left(b+1\right)^{ }\left(c+1\right)^{ }\ge8\) \(\left(đpcm\right)\)
3)
Sửa đề \(A=\dfrac{a}{b+c-a}+\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}\)
Đặt b + c - a = x , a+c-b = y , a+b-c= z
\(\Rightarrow\left[{}\begin{matrix}2a=y+z\\2b=x+z\\2c=x+y\end{matrix}\right.\)
Có :
\(\dfrac{a}{b+c-a}+\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}\)
\(\Rightarrow\dfrac{2a}{b+c-a}+\dfrac{2b}{a+c-b}+\dfrac{2c}{a+b-c}\)
\(\Rightarrow\dfrac{y+z}{x}+\dfrac{x+z}{y}+\dfrac{x+y}{z}\)
\(\Rightarrow\left(\dfrac{y}{x}+\dfrac{x}{y}\right)+\left(\dfrac{z}{x}+\dfrac{x}{z}\right)+\left(\dfrac{z}{y}+\dfrac{y}{z}\right)\)
Áp dụng bất đẳng thức \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\forall a,b>0\)
\(\Rightarrow\) \(\left(\dfrac{y}{x}+\dfrac{x}{y}\right)+\left(\dfrac{z}{x}+\dfrac{x}{z}\right)+\left(\dfrac{z}{y}+\dfrac{y}{z}\right)\ge6\)
\(\Rightarrow\dfrac{2a}{b+c-a}+\dfrac{2b}{a+c-b}+\dfrac{2c}{a+b-c}\ge6\)
\(\Rightarrow2\left(\dfrac{a}{b+c-a}+\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}\right)\ge6\)
\(\Rightarrow\dfrac{a}{b+c-a}+\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}\ge3\) \(\left(đpcm\right)\)
Ta có:\(\dfrac{1}{1+ab}+\dfrac{1}{1+bc}+\dfrac{1}{1+ac}\ge\dfrac{9}{1+1+1+ab+bc+ca}\)(AM-GM)
Lại có:\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ca\)
\(\Rightarrow\dfrac{9}{3+ab+bc+ca}\ge\dfrac{9}{3+a^2+b^2+c^2}=\dfrac{9}{6}=\dfrac{3}{2}\)
\(\Rightarrowđpcm\)
Cháu làm cho bác câu 2 thôi,câu 3 THANGDZ làm rồi sợ mất bản quyền lắm:v
Lời giải:
Áp dụng liên tiếp bất đẳng thức AM-GM và Cauchy-Schwarz ta có:
\(\dfrac{a}{a+2b+3c}+\dfrac{b}{b+2c+3a}+\dfrac{c}{c+2a+3b}\)
\(=\dfrac{a^2}{a^2+2ab+3ac}+\dfrac{b^2}{b^2+2bc+3ab}+\dfrac{c^2}{c^2+2ac+3bc}\)
\(\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+5ab+5bc+5ac}\)
\(=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+3\left(ab+bc+ac\right)}\ge\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\left(a+b+c\right)^2}=\dfrac{1}{2}\)
Ùi mình làm theo kiểu khác thử :V, nhưng có hơi hướng giống và bổ sung :D
Câu 2 : a,b,c > 0. CM : \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\)
Giải :
C1 : Áp dụng bất đẳng thức Cauchy - Schwarz dạng Engel ta có :
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{\left(1+1+1\right)^2}{a+b+c}=\dfrac{9}{a+b+c}\left(ĐPCM\right)\)
Đẳng thức xảy ra khi \(\dfrac{1}{a}=\dfrac{1}{b}=\dfrac{1}{c}\).
C2 : Đầy đủ hơn với cách giải đúng của bạn Hoàng Thiên Di :
Áp dụng BĐT AM-GM cho 3 số dương (sgk là cosi :v)
\(a+b+c\ge3\sqrt[3]{abc}\)\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\sqrt[3]{\dfrac{1}{abc}}\)
\(\Leftrightarrow\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1+1+1+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\)
\(\ge3+2+2+2=9\left(ĐPCM\right)\)
Câu 3 : a,b,c > 0. CM : \(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\ge6\)
Giải :
\(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\ge6\)
\(\Leftrightarrow\dfrac{a}{c}+\dfrac{b}{c}+\dfrac{b}{a}+\dfrac{c}{a}+\dfrac{c}{b}+\dfrac{a}{b}\ge6\)
\(\Leftrightarrow\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\ge6\)
Theo bất đẳng thức Cosi : \(\dfrac{x}{y}+\dfrac{y}{x}\ge2\sqrt{\dfrac{xy}{yx}}=2\)
Thay vào các vế được : \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{ab}{ba}}=2\sqrt{1}=2\)
\(\dfrac{a}{c}+\dfrac{c}{a}\ge2\sqrt{\dfrac{ac}{ca}}=2\sqrt{1}=2\)
\(\dfrac{b}{c}+\dfrac{c}{b}\ge2\sqrt{\dfrac{bc}{cb}}=2\sqrt{1}=2\)
\(\Leftrightarrow2+2+2\ge6\) (đúng)
BĐT được c/m.
b. Áp dụng BĐT Cauchy-Schwarz dạng Engel, ta có:
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}=\dfrac{\left(a+b+c\right)^2}{b+c+a+c+a+b}=\dfrac{\left(a+b+c\right)^2}{2a+2b+2c}=\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\)
a) Áp dụng BĐT Cauchy :
Ta có: \(x+\dfrac{1}{x}\) = \(\dfrac{x^2+1}{x}\) \(\ge\) \(\dfrac{2x}{x}\) = 2 => đpcm
=> Dấu = xảy ra khi x = 1
b) Áp dụng BĐT Svac-sơ ta có:
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}\) \(\ge\) \(\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\)
=> đpcm
=> Dấu bằng xảy ra <=> a = b = c
2a)
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\forall a,b>0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2a+b+c}=\dfrac{1}{a+b+a+c}\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)\\\dfrac{1}{a+2b+c}=\dfrac{1}{a+b+b+c}\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)\\\dfrac{1}{a+b+2c}=\dfrac{1}{a+c+b+c}\le\dfrac{1}{4}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)\end{matrix}\right.\)
\(\Rightarrow VT\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)+\dfrac{1}{4}\left(\dfrac{1}{b+c}+\dfrac{1}{a+b}\right)+\dfrac{1}{4}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)\)
\(\Rightarrow VT\le\dfrac{1}{4\left(a+b\right)}+\dfrac{1}{4\left(a+c\right)}+\dfrac{1}{4\left(b+c\right)}+\dfrac{1}{4\left(a+b\right)}+\dfrac{1}{4\left(a+c\right)}+\dfrac{1}{4\left(b+c\right)}\)
\(\Rightarrow VT\le\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\)
Chứng minh rằng \(\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Leftrightarrow\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\forall a,b>0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\\\dfrac{1}{b+c}\le\dfrac{1}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\\\dfrac{1}{c+a}\le\dfrac{1}{4}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\le\dfrac{1}{4}\left(\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\right)\)
\(\Rightarrow\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) ( đpcm )
Vì \(\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
Mà \(VT\le\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\)
\(\Rightarrow\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)( đpcm )
Dấu " = " xảy ra khi \(a=b=c\)
2b)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}1+a^2\ge2\sqrt{a^2}=2a\\1+b^2\ge2\sqrt{b^2}=2b\\1+c^2\ge2\sqrt{c^2}=2c\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{1+a^2}\le\dfrac{a}{2a}=\dfrac{1}{2}\\\dfrac{b}{1+b^2}\le\dfrac{b}{2b}=\dfrac{1}{2}\\\dfrac{c}{1+c^2}\le\dfrac{c}{2c}=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\dfrac{a}{1+a^2}+\dfrac{b}{1+b^2}+\dfrac{c}{1+c^2}\le\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{3}{2}\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c=1\)
Bài 1)
Nháp : nhìn nhanh ta thấy nên áp dụng BĐT \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)
Giải
Vì x,y > 0 =) 2x + y > 0 , x + 2y > 0
Áp dụng BĐT cauchy dạng phân thức cho hai bộ số không âm \(\dfrac{1}{2x+y}\)và\(\dfrac{1}{x+2y}\)
\(\Rightarrow\dfrac{1}{x+2y}+\dfrac{1}{2x+y}\ge\dfrac{4}{x+2y+2x+y}=\dfrac{4}{3\left(x+y\right)}\)
\(\Rightarrow\left(3x+3y\right)\left(\dfrac{1}{2x+y}+\dfrac{1}{x+2y}\right)\ge\left(3x+3y\right).\dfrac{4}{3\left(x+y\right)}=4\)
Dấu '' = "xảy ra khi và chỉ khi x + 2y = y + 2x (=) x=y
Bài 3:
a) Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{xy}+\frac{2}{x^2+y^2}=2\left(\frac{1}{2xy}+\frac{1}{x^2+y^2}\right)\) \(\geq 2.\frac{(1+1)^2}{2xy+x^2+y^2}=\frac{8}{(x+y)^2}=8\)
Dấu bằng xảy ra khi \(x=y=\frac{1}{2}\)
b) Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{xy}+\frac{1}{x^2+y^2}=\frac{1}{2xy}+\left (\frac{1}{2xy}+\frac{1}{x^2+y^2}\right)\geq \frac{1}{2xy}+\frac{(1+1)^2}{2xy+x^2+y^2}\)
\(=\frac{1}{2xy}+\frac{4}{(x+y)^2}\)
Theo BĐT AM-GM:
\(xy\leq \frac{(x+y)^2}{4}=\frac{1}{4}\Rightarrow \frac{1}{2xy}\geq 2\)
Do đó \(\frac{1}{xy}+\frac{1}{x^2+y^2}\geq 2+4=6\)
Dấu bằng xảy ra khi \(x=y=\frac{1}{2}\)
Bài 1: Thiếu đề.
Bài 2: Sai đề, thử với \(x=\frac{1}{6}\)
Bài 4 a) Sai đề với \(x<0\)
b) Áp dụng BĐT AM-GM:
\(x^4-x+\frac{1}{2}=\left (x^4+\frac{1}{4}\right)-x+\frac{1}{4}\geq x^2-x+\frac{1}{4}=(x-\frac{1}{2})^2\geq 0\)
Dấu bằng xảy ra khi \(\left\{\begin{matrix} x^4=\frac{1}{4}\\ x=\frac{1}{2}\end{matrix}\right.\) (vô lý)
Do đó dấu bằng không xảy ra , nên \(x^4-x+\frac{1}{2}>0\)
Bài 6: Áp dụng BĐT AM-GM cho $6$ số:
\(a^2+b^2+c^2+d^2+ab+cd\geq 6\sqrt[6]{a^3b^3c^3d^3}=6\)
Do đó ta có đpcm
Dấu bằng xảy ra khi \(a=b=c=d=1\)
5) a) Đặt b+c-a=x;a+c-b=y;a+b-c=z thì 2a=y+z;2b=x+z;2c=x+y
Ta có:
\(\dfrac{2a}{b+c-a}+\dfrac{2b}{a+c-b}+\dfrac{2c}{a+b-c}=\dfrac{y+z}{x}+\dfrac{x+z}{y}+\dfrac{x+y}{z}=\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+\left(\dfrac{z}{x}+\dfrac{x}{z}\right)+\left(\dfrac{z}{y}+\dfrac{y}{z}\right)\ge6\)
Vậy ta suy ra đpcm
b) Ta có: a+b>c;b+c>a;a+c>b
Xét: \(\dfrac{1}{a+c}+\dfrac{1}{b+c}>\dfrac{1}{a+b+c}+\dfrac{1}{b+c+a}=\dfrac{2}{a+b+c}>\dfrac{2}{a+b+a+b}=\dfrac{1}{a+b}\)
.Tương tự:
\(\dfrac{1}{a+b}+\dfrac{1}{a+c}>\dfrac{1}{b+c};\dfrac{1}{a+b}+\dfrac{1}{b+c}>\dfrac{1}{a+c}\)
Vậy ta có đpcm
6) Ta có:
\(a^2+b^2+c^2+d^2+ab+cd\ge2ab+2cd+ab+cd=3\left(ab+cd\right)\)
\(ab+cd=ab+\dfrac{1}{ab}\ge2\)
Suy ra đpcm
Lời giải:
Áp dụng BĐT AM-GM: \(a^2+b^2\geq 2ab\Rightarrow 2(a^2+b^2)\geq (a+b)^2\)
\(a+b=a^2+b^2\geq \frac{(a+b)^2}{2}\Rightarrow 2(a+b)\geq (a+b)^2\)
Do đó mà \(a+b\leq 2\)
Có:
\(S=\frac{a}{a+1}+\frac{b}{b+1}=1-\frac{1}{a+1}+1-\frac{1}{b+1}=2-\left(\frac{1}{a+1}+\frac{1}{b+1}\right)\)
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{a+1}+\frac{1}{b+1}\geq \frac{4}{a+b+2}\geq \frac{4}{2+2}=1\) do \(a+b\leq 2\)
Do đó: \(S\leq 2-1\Leftrightarrow S\leq 1\)
Vậy \(S_{\max}=1\Leftrightarrow a=b=1\)
\(S=\dfrac{a}{a+1}+\dfrac{b}{b+1}=\dfrac{a^2}{a^2+a}+\dfrac{b^2}{b^2+b}\)
Áp dụng BĐT Cauchy-Schwarz, ta có:
\(\dfrac{a^2}{a^2+a}+\dfrac{b^2}{b^2+b}\ge\dfrac{\left(a+b\right)^2}{a^2+b^2+a+b}=\dfrac{\left(a+b\right)^2}{2\left(a^2+b^2\right)}\)
Bởi vì:\(a^2+b^2=a+b\Rightarrow a^2+b^2+a+b=a^2+b^2+a^2+b^2=2\left(a^2+b^2\right)\)
Mặt khác, theo Bunyakovsky, ta có:
\(2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)
Do đó: \(\dfrac{\left(a+b\right)^2}{2\left(a^2+b^2\right)}\le\dfrac{\left(a+b\right)^2}{\left(a+b\right)^2}=1\)
Vậy: \(Max_S=1\Leftrightarrow a=b=1\)