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xin lỗi nha MÌNH sai đề ở chổ \(\sqrt{a}+\sqrt{b}+\sqrt{c}=3\)
\(3a^2+8b^2+14ab\le3a^2+8b^2+12ab+a^2+b^2=\left(2a+3b\right)^2\)
\(\Rightarrow\sqrt{3a^2+8b^2+14ab}\le2a+3b\)
\(\Rightarrow P=\sum\frac{a^2}{\sqrt{3a^2+8b^2+14ab}}\ge\sum\frac{a^2}{2a+3b}\ge\frac{\left(a+b+c\right)^2}{5\left(a+b+c\right)}=\frac{a+b+c}{5}\)
Dấu "=" xảy ra khi \(a=b=c\)
Ta có: \(\sqrt{3a^2+14ab+8b^2}=\sqrt{\left(2a+3b\right)^2-\left(a-b\right)^2}\)
\(\le\sqrt{\left(2a+3b\right)^2}=2a+3b\)
Tương tự, ta có: \(\sqrt{3b^2+14bc+8c^2}\le2b+3c\); \(\sqrt{3c^2+14ca+8a^2}\le2c+3a\)
\(\Rightarrow\frac{a^2}{\sqrt{3a^2+14ab+8b^2}}+\frac{b^2}{\sqrt{3b^2+14bc+8c^2}}+\frac{c^2}{\sqrt{3c^2+14ca+8a^2}}\)
\(\ge\frac{a^2}{2a+3b}+\frac{b^2}{2b+3c}+\frac{c^2}{2c+3a}\ge\frac{\left(a+b+c\right)^2}{5\left(a+b+c\right)}=\frac{a+b+c}{5}\)(Theo BĐT Bunyakovski dạng phân thức)
Đẳng thức xảy ra khi a = b = c
b) Ta có:
\(\frac{a}{\sqrt{b^2+3}}+\frac{a}{\sqrt{b^2+3}}+\frac{b^2+3}{8}+\frac{a^2}{2}\)\(\ge\)\(4\sqrt[4]{\frac{a^4}{16}}=2a\)
\(\frac{b}{\sqrt{c^2+3}}+\frac{b}{\sqrt{c^2+3}}+\frac{c^2+3}{8}+\frac{b^2}{2}\ge4\sqrt[4]{\frac{b^4}{16}}=2b\)
\(\frac{c}{\sqrt{a^2+3}}+\frac{c}{\sqrt{a^2+3}}+\frac{a^2+3}{8}+\frac{c^2}{2}\ge4\sqrt[4]{\frac{c^4}{16}}=2c\)
Cộng lại ta đươc:
\(2\left(\frac{a}{\sqrt{b^2+3}}+\frac{b}{\sqrt{c^2+3}}+\frac{c}{\sqrt{a^2+3}}\right)+\)\(\frac{5\left(a^2+b^2+c^2\right)+9}{8}\)\(\ge2\left(a+b+c\right)\)
⇒ \(2\left(\frac{a}{\sqrt{b^2+3}}+\frac{b}{\sqrt{c^2+3}}+\frac{c}{\sqrt{a^2+3}}\right)\ge\)\(6-\frac{5\left(a^2+b^2+c^2\right)+9}{8}\)(1)
Lại có: \(a^2+1\ge2a\); \(b^2+1\ge2b\); \(c^2+1\ge2c\)
Suy ra \(a^2+b^2+c^2\ge2\left(a+b+c\right)-3=3\)
Khi đó (1)⇔ \(2\left(\frac{a}{\sqrt{b^2+3}}+\frac{b}{\sqrt{c^2+3}}+\frac{c}{\sqrt{a^2+3}}\right)\ge\)\(6-\frac{5.3+9}{8}=3\)
⇒ \(\frac{a}{\sqrt{b^2+3}}+\frac{b}{\sqrt{c^2+3}}+\frac{c}{\sqrt{a^2+3}}\ge\frac{3}{2}\)
Dấu "=" xảy ra ⇔ \(a=b=c=1\)
\(\left(a^2+3b^2\right)\left(1+3\right)\ge\left(a+3b\right)^2\Rightarrow\sqrt{a^2+3b^2}\ge\frac{a+3b}{2}\)
\(\Rightarrow P=\sum\frac{ab}{\sqrt{a^2+3b^2}}\le2\sum\frac{ab}{a+3b}=2\sum\frac{ab}{a+b+b+b}\)
\(\Rightarrow P\le\frac{1}{8}\sum ab\left(\frac{1}{a}+\frac{3}{b}\right)=\frac{1}{8}\sum\left(3a+b\right)=\frac{1}{2}\left(a+b+c\right)=\frac{3}{2}\)
"=" \(\Leftrightarrow a=b=c=1\)
Ta có: \(\sqrt{3a^2+8b^2+14ab}=\sqrt{\left(3a+2b\right)\left(a+4b\right)}\le2a+3b\)
Khi đó \(\frac{a^2}{\sqrt{3a^2+8b^2+14ab}}\ge\frac{a^2}{2a+3b}\), tương tự cho ta cũng có:
\(\frac{b^2}{\sqrt{3b^2+8c^2+14bc}}\ge\frac{b^2}{2b+3c};\frac{c^2}{\sqrt{3c^2+8a^2+14ca}}\ge\frac{c^2}{2c+3a}\)
Cộng theo vế ta có: \(\frac{a^2}{\sqrt{3a^2+8b^2+14ab}}+\frac{b^2}{\sqrt{3b^2+8c^2+14bc}}+\frac{c^2}{\sqrt{3c^2+8a^2+14ca}}\)
\(\ge\frac{a^2}{2a+3b}+\frac{b^2}{2b+3c}+\frac{c^2}{2c+3a}\ge\frac{\left(a+b+c\right)^2}{5\left(a+b+c\right)}=\frac{a+b+c}{5}\)
\(\frac{a^2}{\sqrt{3a^2+8b^2+14ab}}+\frac{b^2}{\sqrt{3b^2+8c^2+14bc}}+\frac{c^2}{\sqrt{3c^2+8a^2+14ca}}\)
\(\Leftrightarrow\frac{a^2}{\sqrt{3a^2+12ab+8b^2+2ab}}+\frac{b^2}{\sqrt{3b^2+12bc+8c^2+2bc}}+\frac{c^2}{\sqrt{3c^2+12ca+8a^2+2ca}}\)
\(\Leftrightarrow\frac{a^2}{\sqrt{3a\left(a+4b\right)+2b\left(4b+a\right)}}+\frac{b^2}{\sqrt{3b\left(b+4c\right)+2c\left(4c+b\right)}}+\frac{c^2}{\sqrt{3c\left(c+4a\right)+2a\left(4a+c\right)}}\)
\(\Leftrightarrow\frac{a^2}{\sqrt{\left(a+4b\right)\left(3a+2b\right)}}+\frac{b^2}{\sqrt{\left(b+4c\right)\left(3b+2c\right)}}+\frac{c^2}{\sqrt{\left(c+4a\right)\left(3c+2a\right)}}\)
Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm
\(\Rightarrow\left\{\begin{matrix}\sqrt{\left(a+4b\right)\left(3a+2b\right)}\le\frac{4a+6b}{2}\\\sqrt{\left(b+4c\right)\left(3b+2c\right)}\le\frac{4b+6c}{2}\\\sqrt{\left(c+4a\right)\left(3c+2a\right)}\le\frac{4c+6a}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}\frac{a^2}{\sqrt{\left(a+4b\right)\left(3a+2b\right)}}\ge\frac{2a^2}{4a+6b}\\\frac{b^2}{\sqrt{\left(b+4c\right)\left(3b+2c\right)}}\ge\frac{2b^2}{4b+6c}\\\frac{c^2}{\sqrt{\left(c+4a\right)\left(3c+2a\right)}}\ge\frac{2c^2}{4c+6a}\end{matrix}\right.\)
\(\Rightarrow VT\ge\frac{2a^2}{4a+6b}+\frac{2b^2}{4b+6c}+\frac{2c^2}{4c+6a}\)
Chứng minh rằng \(\frac{2a^2}{4a+6b}+\frac{2b^2}{4b+6c}+\frac{2c^2}{4c+6a}\ge\frac{1}{5}\left(a+b+c\right)\)
\(\Leftrightarrow2\left(\frac{a^2}{4a+6b}+\frac{b^2}{4b+6c}+\frac{c^2}{4c+6a}\right)\ge\frac{1}{5}\left(a+b+c\right)\)
Áp dụng bất đẳng thức cộng mẫu số
\(\Rightarrow\frac{a^2}{4a+6b}+\frac{b^2}{4b+6c}+\frac{c^2}{4c+6a}\ge\frac{\left(a+b+c\right)^2}{10\left(a+b+c\right)}\)
\(\Rightarrow2\left(\frac{a^2}{4a+6b}+\frac{b^2}{4b+6c}+\frac{c^2}{4c+6a}\right)\ge\frac{2\left(a+b+c\right)^2}{10\left(a+b+c\right)}=\frac{a+b+c}{5}\)
\(\Rightarrow2\left(\frac{a^2}{4a+6b}+\frac{b^2}{4b+6c}+\frac{c^2}{4c+6a}\right)\ge\frac{1}{5}\left(a+b+c\right)\)
Vậy \(\frac{2a^2}{4a+6b}+\frac{2b^2}{4b+6c}+\frac{2c^2}{4c+6a}\ge\frac{1}{5}\left(a+b+c\right)\)
Mà \(VT\ge\frac{2a^2}{4a+6b}+\frac{2b^2}{4b+6c}+\frac{2c^2}{4c+6a}\)
\(\Rightarrow VT\ge\frac{1}{5}\left(a+b+c\right)\)
\(\Leftrightarrow\frac{a^2}{\sqrt{3a^2+8b^2+14ab}}+\frac{b^2}{\sqrt{3b^2+8c^2+14bc}}+\frac{c^2}{\sqrt{3c^2+8a^2+14ca}}\ge\frac{1}{5}\left(a+b+c\right)\)
( đpcm )
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)+abc\)
\(=abc+a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+abc+abc\)
\(=\left(a+b+c\right)\left(ab+bc+ca\right)\)( phân tích nhân tử các kiểu )
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\left(a+b+c\right)\left(ab+bc+ca\right)-abc\left(1\right)\)
\(a+b+c\ge3\sqrt[3]{abc};ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\Rightarrow\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc\)
\(\Rightarrow-abc\ge\frac{-\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)
Khi đó:\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(\ge\left(a+b+c\right)\left(ab+bc+ca\right)-\frac{\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)
\(=\frac{8\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\left(2\right)\)
Từ ( 1 ) và ( 2 ) có đpcm
Chú ý: \(\left(a^2+2b^2+c^2\right)\left(2^2+1^2+2^2\right)\ge\left(2a+2b+2c\right)^2\)
\(\Rightarrow a^2+2b^2+c^2\ge\frac{4\left(a+b+c\right)^2}{9}\Rightarrow\sqrt{a^2+2b^2+c^2}\ge\frac{2}{3}\left(a+b+c\right)\)
Tương tự: \(\sqrt{b^2+2c^2+a^2}\ge\frac{2}{3}\left(a+b+c\right)\); \(\sqrt{c^2+2a^2+b^2}\ge\frac{2}{3}\left(a+b+c\right)\)
Thay vào ta có: \(VT\le\frac{3\left(3a+b+3b+c+3c+a\right)}{2\left(a+b+c\right)}=6\)(qed)
Đẳng thức xảy ra khi a = b = c
Is that true?
Áp dụng bđt Bunhiacopxki ta được:
\(\left(\text{Σ}_{cyc}\frac{3a+b}{\sqrt{a^2+2b^2+c^2}}\right)^2\le3\left(\text{Σ}_{cyc}\frac{\left(3a+b\right)^2}{a^2+2b^2+c^2}\right)\)
Mặt khác cũng theo bđt Bunhiacopxki dạng phân thức, ta được:
\(\frac{\left(3a+b\right)^2}{a^2+2b^2+c^2}\le\frac{9a^2}{a^2+b^2+c^2}+\frac{b^2}{b^2}=\frac{9a^2}{a^2+b^2+c^2}+1\)
Hoàn toàn tương tự, ta có:
\(\frac{\left(3b+c\right)^2}{b^2+2c^2+a^2}\le\frac{9b^2}{b^2+c^2+a^2}+1\);\(\frac{\left(3c+a\right)^2}{c^2+2a^2+b^2}\le\frac{9c^2}{c^2+a^2+b^2}+1\)
Cộng từng vế của các bđt trên, ta được:
\(\text{}\text{}\text{Σ}_{cyc}\frac{\left(3b+c\right)^2}{b^2+2c^2+a^2}\le\text{Σ}_{cyc}\frac{9b^2}{b^2+c^2+a^2}+3=9+3=12\)
Do đó \(\left(\text{Σ}_{cyc}\frac{3a+b}{\sqrt{a^2+2b^2+c^2}}\right)^2\le3\left(\text{Σ}_{cyc}\frac{\left(3a+b\right)^2}{a^2+2b^2+c^2}\right)\le3.12=36\)
Hay \(\left(\text{Σ}_{cyc}\frac{3a+b}{\sqrt{a^2+2b^2+c^2}}\right)\le6\)
Đẳng thức xảy ra khi a = b = c
Theo BĐT Bu nhi a cốp xki ta có :
\(VT=\sqrt{a+3b}+\sqrt{b+3c}+\sqrt{c+3a}\le\sqrt{3\left(4a+4b+4c\right)}=\sqrt{12\left(a+b+c\right)}=\sqrt{36}=6\)
Vậy đpcm . Dấu bằng xảy ra khi \(a=b=c=1\)