xí câu 1:))

Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel ta có :

\(\frac{x^2}{y-1}+\frac{y^2}{x-1}\ge\frac{\left(x+y\right)^2}{x+y-2}\)(1)

Đặt a = x + y - 2 => a > 0 ( vì x,y > 1 )

Khi đó \(\left(1\right)=\frac{\left(a+2\right)^2}{a}=\frac{a^2+4a+4}{a}=\left(a+\frac{4}{a}\right)+4\ge2\sqrt{a\cdot\frac{4}{a}}+4=8\)( AM-GM )

Vậy ta có đpcm

Đẳng thức xảy ra <=> a=2 => x=y=2

a. Đề bài sai (thực chất là nó đúng 1 cách hiển nhiên nhưng "dạng" thế này nó sai sai vì ko ai cho kiểu này cả)

Ta có: \(abc=ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\Rightarrow abc\ge27\)

\(\Rightarrow a^2+b^2+c^2+5abc\ge a^2+b^2+c^2+5.27>>>>>8\)

b. 

\(4=ab+bc+ca+abc=ab+bc+ca+\sqrt{ab.bc.ca}\le ab+bc+ca+\sqrt{\left(\dfrac{ab+bc+ca}{3}\right)^3}\)

\(\sqrt{\dfrac{ab+bc+ca}{3}}=t\Rightarrow t^3+3t^2-4\ge0\Rightarrow\left(t-1\right)\left(t+2\right)^2\ge0\)

\(\Rightarrow t\ge1\Rightarrow ab+bc+ca\ge3\Rightarrow a+b+c\ge\sqrt{3\left(ab+bc+ca\right)}\ge3\)

- TH1: nếu \(a+b+c\ge4\)

Ta có: \(ab+bc+ca=4-abc\le4\)

\(\Rightarrow P=\left(a+b+c\right)^2-2\left(ab+bc+ca\right)+5abc\ge4^2-2.4+0=8\)

(Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(2;2;0\right)\) và các hoán vị)

- TH2: nếu \(3\le a+b+c< 4\)

Đặt \(a+b+c=p\ge3;ab+bc+ca=q;abc=r\)

\(P=p^2-2q+5r=p^2-2q+5\left(4-q\right)=p^2-7q+20\)

Áp dụng BĐT Schur:

\(4=q+r\ge q+\dfrac{p\left(4q-p^2\right)}{9}\Leftrightarrow q\le\dfrac{p^3+36}{4p+9}\)

\(\Rightarrow P\ge p^2-\dfrac{7\left(p^3+36\right)}{4p+9}+20=\dfrac{3\left(4-p\right)\left(p-3\right)\left(p+4\right)}{4p+9}+8\ge8\)

(Dấu "=" xảy ra khi \(a=b=c=1\))

1) \(\Sigma\frac{a}{b^3+ab}=\Sigma\left(\frac{1}{b}-\frac{b}{a+b^2}\right)\ge\Sigma\frac{1}{a}-\Sigma\frac{1}{2\sqrt{a}}=\Sigma\left(\frac{1}{a}-\frac{2}{\sqrt{a}}+1\right)+\Sigma\frac{3}{2\sqrt{a}}-3\)

\(\ge\Sigma\left(\frac{1}{\sqrt{a}}-1\right)^2+\frac{27}{2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}-3\ge\frac{27}{2\sqrt{3\left(a+b+c\right)}}-3=\frac{3}{2}\)

2.

Vỉ \(ab+bc+ca+abc=4\)thi luon ton tai \(a=\frac{2x}{y+z};b=\frac{2y}{z+x};c=\frac{2z}{x+y}\)

\(\Rightarrow VT=2\Sigma_{cyc}\sqrt{\frac{ab}{\left(b+c\right)\left(c+a\right)}}\le2\Sigma_{cyc}\frac{\frac{b}{b+c}+\frac{a}{c+a}}{2}=3\)

Cho o dong 2 la x,y,z nhe,ghi nham

\(a^2+b^2+c^2+2ab-2ac-2bc=a^2+b^2\)

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

\(\Rightarrow\hept{\begin{cases}a^2=\left(a+b-c\right)^2-b^2=\left(a+b-c-b\right)\left(a+b-c+b\right)=\left(a-c\right)\left(a+2b-c\right)\\b^2=\left(a+b-c\right)^2-a^2=\left(a+b-c-a\right)\left(a+b-c+a\right)=\left(b-c\right)\left(2a+b-c\right)\end{cases}}\)

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

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

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

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

Vậy \(\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\frac{2\left(a-c\right)\left(a+b+c\right)}{2\left(b-c\right)\left(a+b+c\right)}=\frac{a-c}{b-c}\)

a)a²+b²+c²+3=2(a+b+c)

=>a²+b²+c²+1+1+1-2a-2b-2c=0

=>(a²-2a+1)+(b²-2b+1)+(c²-2c+1)=0

=>(a-1)²+(b-1)²+(c-1)²=0

=>a-1=b-1=c-1=0 <=>a=b=c=1 

-->Đpcm

b)(a+b+c)²=3(ab+ac+bc)

=>a²+b²+c²+2ab+2ac+2bc -3ab-3ac-3bc=0 

=>a²+b²+c²-ab-ac-bc=0

=>2a²+2b²+2c²-2ab-2ac-2bc=0 

=>(a²- 2ab+b²)+(b²-2bc+c²) + (c²-2ca+a²) = 0

=>(a-b)²+(b-c)²+(c-a)²=0 

Hay (a-b)²=0 hoặc (b-c)²=0 hoặc (a-c)²=0

=>a-b hoặc b=c hoặc a=c

=>a=b=c 

-->Đpcm

c)a²+b²+c²=ab+bc+ca

=>2(a²+b²+c²)=2(ab+bc+ca)

=>2a²+2b²+c²=2ab+2bc+2ca

=>2a²+2b²+c²-2ab-2bc-2ca=0

=>a²+a²+b²+b²+c²+c²-2ab-2bc-2ca=0

=>(a²-2ab+b²)+(b²-2bc+c²)+(a²-2ca+c²)=0

=>(a-b)²+(b-c)²+(a-c)²=0

Hay (a-b)²=0 hoặc (b-c)²=0 hoặc (a-c)²=0

=>a-b hoặc b=c hoặc a=c

=>a=b=c 

-->Đpcm

a) Ta có : \(a^2+b^2+c^2+3=2\left(a+b+c\right)\)

\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)=0\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)

Vì \(\left(a-1\right)^2\ge0,\left(b-1\right)^2\ge0,\left(c-1\right)^2\ge0\) nên pt trên tương đương với \(\begin{cases}\left(a-1\right)^2=0\\\left(b-1\right)^2=0\\\left(c-1\right)^2=0\end{cases}\) \(\Leftrightarrow a=b=c=1\)

b) \(\left(a+b+c\right)^2=3\left(ab+bc+ac\right)\)

\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=3\left(ab+bc+ac\right)\)

\(\Leftrightarrow a^2+b^2+c^2=ab+bc+ac\) (1)

\(\Leftrightarrow2\left(a^2+b^2+c^2\right)=2ab+2bc+2ac\)

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

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

Mà \(\left(a-b\right)^2\ge0,\left(b-c\right)^2\ge0,\left(c-a\right)^2\ge0\)

\(\Rightarrow\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}\) \(\Rightarrow a=b=c\)

c) Giải tương tự câu b) , bắt đầu từ (1)