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Ta có : \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\Leftrightarrow\left(3+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)< 10\)
\(\Leftrightarrow\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{b}{a}+\frac{c}{a}+\frac{a}{c}< 7\)
\(\Leftrightarrow\frac{a+c}{b}+\frac{b+a}{c}+\frac{c+b}{a}< 7\)
Không giảm tổng quá .Giả sử a là cạnh lớn nhất .Giả b + c < a => 0 < \(\frac{b+c}{a}\)
\(\Rightarrow\frac{a+c}{b}+\frac{b+a}{c}+\frac{c+b}{a}>\frac{2c+b}{b}+\frac{2b+c}{c}+\frac{b+c}{a}\)( không chắc lắm )
= \(\frac{2c}{b}+\frac{2b}{c}+\frac{b+c}{a}+2\)
=\(\frac{2\left(b+c\right)^2}{bc}+\frac{b+c}{a}-2>7\left(VL\right)\)
=>b+ c > a => a ; b ; c là 3 cạnh tam giác ( đpcm )
Bạn kia làm sai r
Ta có đánh giá quen thuộc \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=\frac{a+b+c}{abc}\)
mà \(3abc\left(a+b+c\right)\le\left(ab+bc+ca\right)^2\)
do đó \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{a+b+c}{abc}=\frac{\left(a+b+c\right)^2}{abc\left(a+b+c\right)}\ge\frac{3\left(a+b+c\right)^2}{\left(ab+bc+ca\right)^2}\)
Phép chứng minh hoàn tất khi ta cm được
\(\frac{3\left(a+b+c\right)^2}{\left(ab+bc+ca\right)^2}\ge a^2+b^2+c^2\)
hay \(3\left(a+b+c\right)^2\ge\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)^2\)
Theo bđt AM-GM ta có
\(\left(a+b+c\right)^2=\left(a^2+b^2+c^2\right)+\left(ab+bc+ca\right)+\left(ab+bc+ca\right)\)
\(\ge3\sqrt[3]{\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)^2}\)
hay \(\left(a+b+c\right)^6\ge27\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)^2\)
mà a+b+c=3 nên \(\left(a+b+c\right)^6=81\left(a+b+c\right)^2\)
\(\Rightarrow3\left(a+b+c\right)^2\ge\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)^2\)
Vậy bđt được chứng minh
Dấu "=" xảy ra khi a=b=c=1
Xét BĐT phụ \(\frac{1}{a^2}+4a\ge a^2+4\Leftrightarrow\frac{\left(a-1\right)^2\left(1+2a-a^2\right)}{a^2}\ge0\)
Đến đây, ta đưa điều phải chứng minh về dạng \(\frac{\left(a-1\right)^2\left(1+2a-a^2\right)}{a^2}+\frac{\left(b-1\right)^2\left(1+2b-b^2\right)}{b^2}+\frac{\left(c-1\right)^2\left(1+2c-c^2\right)}{c^2}\ge0\)(*)
Không mất tính tổng quát, giả sử \(a\ge b\ge c\)
Xét hai trường hợp:
Trường hợp 1: \(a\le1+\sqrt{2}\Rightarrow c\le b\le a\le1+\sqrt{2}\)
Khi đó thì \(1+2a-a^2\ge0;1+2b-b^2\ge0;1+2c-c^2\ge0\)dẫn đến (*) đúng
Trường hợp 2: \(a>1+\sqrt{2}\Rightarrow b+c=3-a< 3-\left(1+\sqrt{2}\right)=2-\sqrt{2}< \frac{2}{3}\)
\(\Rightarrow bc\le\frac{\left(b+c\right)^2}{4}< \frac{\frac{4}{9}}{4}=\frac{1}{9}\)
Mà a,b,c dương nên \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}>\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{2}{bc}>18>\left(a+b+c\right)^2>a^2+b^2+c^2\)
Vậy bất đẳng thức được chứng minh
Đẳng thức xảy ra khi a = b = c = 1
Ta có : \(a^2+2b+3=a^2+1+2b+2\ge2a+2b+2=2\left(a+c+1\right)\)
\(b^2+2c+3=b^2+1+2c+2\ge2b+2c+2=2\left(b+c+1\right)\)
\(c^2+2a+3=c^2+1+2a+2\ge2c+2a+2=2\left(c+a+1\right)\)
Suy ra \(\frac{a}{a^2+2b+3}+\frac{b}{b^2+2c+3}+\frac{c}{c^2+2a+3}\le\frac{a}{2\left(a+b+1\right)}+\frac{b}{2\left(b+c+1\right)}+\frac{c}{2\left(c+a+1\right)}\)
\(=\frac{1}{2}\left(\frac{a}{a+b+1}+\frac{b}{b+c+1}+\frac{c}{c+a+1}\right)\)
Tương đương \(\frac{3}{2}-\frac{a}{a^2+2b+3}-\frac{b}{b^2+2c+3}-\frac{c}{c^2+2a+3}\ge\frac{1}{2}\left(\frac{b+1}{a+b+1}+\frac{c+1}{b+c+1}+\frac{a+1}{c+a+1}\right)\)
Đặt \(M=\frac{b+1}{a+b+1}+\frac{c+1}{b+c+1}+\frac{a+1}{c+a+1}\)
Áp dụng bất đẳng thức Cauchy-Schwarz ta được : \(M=\frac{\left(b+1\right)^2}{\left(b+1\right)\left(a+b+1\right)}+\frac{\left(c+1\right)^2}{\left(c+1\right)\left(b+c+1\right)}+\frac{\left(a+1\right)^2}{\left(a+1\right)\left(c+a+1\right)}\)
\(\ge\frac{\left(a+b+c+3\right)^2}{\left(a+1\right)\left(a+b+1\right)+\left(c+1\right)\left(b+c+1\right)+\left(a+1\right)\left(c+a+1\right)}\)
Do \(\left(a+1\right)\left(a+b+1\right)+\left(c+1\right)\left(b+c+1\right)+\left(a+1\right)\left(c+a+1\right)=a^2+b^2+c^2+ab+bc+ca+3\left(a+b+c\right)+3\)\(=\frac{1}{2}\left(a^2+b^2+c^2\right)+ab+bc+ca+3\left(a+b+c\right)+\frac{9}{2}=\frac{1}{2}\left(a+b+c+3\right)^2\)
Từ đó \(M\ge\frac{\left(a+b+c+3\right)^2}{\frac{1}{2}\left(a+b+c+3\right)^2}=2\Rightarrow\frac{3}{2}-\frac{a}{a^2+2b+3}-\frac{b}{b^2+2c+3}-\frac{c}{c^2+2a+3}\ge\frac{1}{2}.2=1\)
\(< =>\frac{a}{a^2+2b+3}+\frac{b}{b^2+2c+3}+\frac{c}{c^2+2a+3}\le\frac{1}{2}\left(đpcm\right)\)
Bài toán hoàn tất . Đẳng thức xảy ra khi và chỉ khi \(a=b=c=1\)
\(https://scontent.fhph1-1.fna.fbcdn.net/v/t34.0-12/19987311_122536408488931_1351154453_n.jpg?oh=553755e5363013e1853ab6f5ed63a600&oe=59BF5CA7\)https://scontent.fhph1-1.fna.fbcdn.net/v/t34.0-12/19987311_122536408488931_1351154453_n.jpg?oh=553755e5363013e1853ab6f5ed63a600&oe=59BF5CA7
Ấn vào linh đấy ế
Ta có: \(\left(\sqrt{a}+\sqrt{c}\right)^2=a+2\sqrt{ac}+c=2b+2\sqrt{ac}\)(1)
Lại có: \(\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{1}{\sqrt{b}+\sqrt{c}}=\frac{2\sqrt{b}+\sqrt{a}+\sqrt{c}}{b+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}\)
\(=\frac{\left(2\sqrt{b}+\sqrt{a}+\sqrt{c}\right)\left(\sqrt{a}+\sqrt{c}\right)}{\left(b+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\left(\sqrt{a}+\sqrt{c}\right)}\)(Nhân cả tử & mẫu với \(\sqrt{a}+\sqrt{c}\))
\(=\frac{2\sqrt{ab}+2\sqrt{bc}+\left(\sqrt{a}+\sqrt{c}\right)^2}{\left(b+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\left(\sqrt{a}+\sqrt{c}\right)}\)(2)
Thế (1) và (2) => \(\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{1}{\sqrt{b}+\sqrt{c}}\)\(=\frac{2\sqrt{ab}+2\sqrt{bc}+2b+\sqrt{ca}}{\left(b+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\left(\sqrt{a}+\sqrt{c}\right)}=\frac{2\left(b+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)}{\left(b+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\left(\sqrt{a}+\sqrt{c}\right)}\)
\(=\frac{2}{\sqrt{a}+\sqrt{c}}.\)
\(\Rightarrow\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{1}{\sqrt{b}+\sqrt{c}}=\frac{2}{\sqrt{a}+\sqrt{c}}\)(đpcm).