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1. Ta có : x + y + z = 0 \(\Rightarrow\)( x + y + z )2 = 0 \(\Rightarrow\)x2 + y2 + z2 = - 2 ( xy + yz + xz )\(S=\frac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}=\frac{-2\left(xy+yz+xz\right)}{2\left(x^2+y^2+z^2\right)-2\left(yz+xz+xy\right)}\)
\(S=\frac{-2\left(xy+yz+xz\right)}{-4\left(xy+yz+xz\right)-2\left(yz+xz+xy\right)}=\frac{-2\left(xy+yz+xz\right)}{-6\left(xy+yz+xz\right)}=\frac{1}{3}\)
Ta có
a + b + c = abc
\(\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Ta lại có
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\)
Ta có:a+b+c=abc
\(\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Ta lại có :\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=4\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\)
\(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac=0\)
\(\Leftrightarrow a^2+b^2-c^2=-2c^2-2bc-2ac-2ab\)
\(\Leftrightarrow a^2+b^2-c^2=-\left[2c.\left(c+b\right)+2a.\left(c+b\right)\right]\)
\(\Leftrightarrow a^2+b^2-c^2=-2.\left(a+c\right)\left(c+b\right)\)
Tương tự \(b^2+c^2-a^2=-2.\left(a+b\right)\left(a+c\right)\)
\(c^2+a^2-b^2=-2.\left(b+c\right)\left(b+a\right)\)
Đặt \(A=\frac{1}{a^2+b^2-c^2}+\frac{1}{b^2+c^2-a^2}+\frac{1}{c^2+a^2-b^2}\)
\(=-\frac{1}{2}.\left[\frac{1}{\left(b+c\right)\left(a+c\right)}+\frac{1}{\left(a+b\right)\left(a+c\right)}+\frac{1}{\left(b+c\right)\left(a+b\right)}\right]\)
\(=-\frac{1}{2}.\frac{a+b+b+c+a+c}{\left(b+c\right).\left(a+c\right)\left(a+b\right)}=-\frac{1}{2}.\frac{2.\left(a+b+c\right)}{\left(b+c\right).\left(a+c\right).\left(a+b\right)}=0\)
\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1\)
\(\left(a+b+c\right)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)=a+b+c\)
\(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}+a+b+c=a+b+c\)
\(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0\)
đpcm
Bài 1:Cách thông thường nhất là sos hoặc cauchy-Schwarz nhưng thôi ko làm:v Thử cách này cho nó mới dù rằng ko chắc
Giả sử \(a\ge b\ge c\Rightarrow c\le1\Rightarrow a+b=3-c\ge2\) và \(a\ge1\)
Ta có \(LHS=a^3.a+b^3.b+c^3.c\)
\(=\left(a^3-b^3\right)a+\left(b^3-c^3\right)\left(a+b\right)+c^3\left(a+b+c\right)\)
\(\ge\left(a^3-b^3\right).1+\left(b^3-c^3\right).2+3c^3\)
\(=a^3+b^3+c^3=RHS\)
Đẳng thức xảy ra khi a = b = c = 1
\(\left(a+b+c\right)^2=a^2+b^2+c^2\)
\(\text{Mà }\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\Rightarrow2ab+2bc+2ac=0\)
\(\Rightarrow\hept{\begin{cases}2ab=-2bc-2ac\\2bc=-2ac-2ab\\2ac=-2ab-2bc\end{cases}}\)
\(A=\frac{a^2}{a^2-2ab-2ac}+\frac{b^2}{b^2-2ab-2bc}+\frac{c^2}{c^2-2bc-2ac}\)
\(A=\frac{a^2}{a.\left(a-2b-2c\right)}+\frac{b^2}{b.\left(b-2a-2c\right)}+\frac{c^2}{c.\left(c-2b-2c\right)}\)
\(A=\frac{a}{a-2b-2c}+\frac{b}{b-2a-2c}+\frac{c}{c-2b-2c}\)
Nhân cả 2 vế với a+b+c
Chứng minh \(\frac{a}{b}+\frac{b}{a}\ge2\) tương tự với \(\frac{b}{c}+\frac{c}{b};\frac{c}{a}+\frac{a}{c}\)
\(\Leftrightarrow\frac{a}{b}+\frac{b}{a}-2\ge0\Leftrightarrow\frac{a^2-2ab+b^2}{ab}\ge0\Leftrightarrow\frac{\left(a-b\right)^2}{ab}\ge0\)luôn đúng do a;b>0
dễ rồi nhé
b) \(P=\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}\)
\(P=\left(\frac{x+1}{x+1}+\frac{y+1}{y+1}+\frac{z+1}{z+1}\right)-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
\(P=\left(1+1+1\right)-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
\(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
Áp dụng bđt Cauchy Schwarz dạng Engel (mình nói bđt như vậy,chỗ này bạn cứ nói theo cái bđt đề bài cho đi) ta được:
\(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\ge\frac{\left(1+1+1\right)^2}{x+1+y+1+z+1}=\frac{9}{4}\)
=>\(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\le3-\frac{9}{4}=\frac{3}{4}\)
=>Pmax=3/4 <=> x=y=z=1/3
\(\Rightarrow\left\{{}\begin{matrix}a=\left(b+c\right)\\b=\left(c-a\right)\\c=\left(a-b\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^2=\left(b+c\right)^2\\b^2=\left(c-a\right)^2\\c^2=\left(a-b\right)^2\end{matrix}\right.\)Thay vào D đc
\(D=\frac{1}{b^2+2bc+c^2+b^2-c^2}+\frac{1}{c^2-2ac+a^2+c^2-a^2}+\frac{1}{a^2-2ab+b^2+a^2-b^2}\)
\(\Leftrightarrow D=\frac{1}{2b\left(b+c\right)}+\frac{1}{2c\left(c-a\right)}+\frac{1}{2a\left(a-b\right)}\)
\(\Leftrightarrow D=\frac{1}{2}\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)
\(\Leftrightarrow D=\frac{1}{2}\left(\frac{a+b+c}{abc}\right)\)
\(\Leftrightarrow D=\frac{1}{2}\left(\frac{b+c+c-a+a-b}{abc}\right)\)
\(\Leftrightarrow D=\frac{1}{2}.\frac{2c}{abc}\)
\(\Leftrightarrow D=\frac{1}{ab}\)
Xem lại xem có sai đề ko, vì mk thường làm bài này với Điều kiện đề là a+b+c=0 chứ chưa từng làm dạng a-b-c=0