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Gọi VT = T
Đặt \(x=3a+b+c;y=3b+c+a;z=3c+a+b\)
\(\Rightarrow x+y+z=5\left(a+b+c\right)=5\left(x-2a\right)=5\left(y-2b\right)\)
\(=5\left(z-2c\right)\)
\(\Rightarrow4x-\left(y+z\right)=10a;4y-\left(z+x\right)=10b;4z-\left(x+y\right)=10c\)
\(\Rightarrow10T=\frac{4x-\left(y+z\right)}{x}+\frac{4y-\left(z+x\right)}{y}+\frac{4z-\left(x+y\right)}{z}\)
\(=12-\left(\frac{y}{x}+\frac{z}{x}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}+\frac{y}{z}\right)\le12-6=6\)
\(\Rightarrow T\le\frac{6}{10}=\frac{3}{5}\)
Dấu "=" khi a = b = c
Đặt \(\hept{\begin{cases}x=3a+b+c\\y=3b+a+c\\z=3c+a+b\end{cases}\left(x;y;z>0\right)}\)
\(\Rightarrow x+y+z=5a+5b+5c=5\left(a+b+c\right)\)
Lại có: \(a+b+c=x-2a=y-2b=z-2c\)
\(\Rightarrow x+y+z=5\left(x-2a\right)=5\left(y-2b\right)=5\left(z-2c\right)\)
\(\Rightarrow4x-\left(y+z\right)=4\left(3a+b+c\right)-\left(4b+4c+2a\right)=10a\)
Tương tự ta có:\(4y-\left(x+z\right)=10b;4z-\left(x+y\right)=10c\)
\(\Rightarrow10T=\frac{4x-\left(y+z\right)}{x}+\frac{4y-\left(x+z\right)}{y}+\frac{4z-\left(x+y\right)}{z}\)
\(=12-\frac{y+z}{x}+\frac{x+z}{y}+\frac{x+y}{z}\)
\(=12-\left(\frac{y}{x}+\frac{z}{x}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}+\frac{y}{z}\right)\)\(\le12-6=6\)(Bđt Cô si)
\(\Rightarrow10T\le6\Rightarrow T\le\frac{6}{10}=\frac{3}{5}\)(Đpcm)
Dấu = khi a=b=c
Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{a}{3a+b+c}=\frac{a}{\frac{a+b+c}{3}+\frac{a+b+c}{3}+\frac{a+b+c}{3}+a+a}\leq \frac{a}{25}\left(\frac{1}{\frac{a+b+c}{3}}+\frac{1}{\frac{a+b+c}{3}}+\frac{1}{\frac{a+b+c}{3}}+\frac{1}{a}+\frac{1}{a}\right)\)
hay \(\frac{a}{3a+b+c}\leq \frac{9a}{25(a+b+c)}+\frac{2}{25}\)
Hoàn toàn TT: \(\frac{b}{a+3b+c}\leq \frac{9b}{25(a+b+c)}+\frac{2}{25}; \frac{c}{a+b+3c}\leq \frac{9c}{25(a+b+c)}+\frac{2}{25}\)
Cộng theo vế các BĐT trên
\(\Rightarrow T\leq \frac{9(a+b+c)}{25(a+b+c)}+\frac{6}{25}=\frac{3}{5}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
Akai Haruma: em có một cách khác là chuẩn hóa, nhưng ko biết đúng không. Vì cô làm cách kia rồi nên em làm cách này, chứ em thích cách kia hơn.
BĐT trên là thuần nhất (đồng bậc) nên chuẩn hóa a + b + c = 3. Ta cần chứng minh:
\(\Sigma\frac{a}{2a+3}\le\frac{3}{5}\)
C1: Áp dụng BđT AM-GM \(\frac{a}{2a+3}=\frac{a}{a+a+1+1+1}\le\left(\frac{1}{25}+\frac{1}{25}+\frac{3a}{25}\right)\)
Tương tự hai BĐT còn lại và cộng theo vế ta thu được đpcm.
Cách 2: (ko hay + dài)
\(BĐT\Leftrightarrow\Sigma\left(\frac{a}{2a+3}-\frac{1}{5}\right)\le0\) \(\Leftrightarrow\Sigma\left(\frac{3\left(a-1\right)}{5\left(2a+3\right)}-\frac{3}{25}\left(a-1\right)\right)+\Sigma\frac{3}{25}\left(a-1\right)\ge0\)
\(\Leftrightarrow\Sigma\left(a-1\right)\left(\frac{3}{5\left(2a+3\right)}-\frac{3}{25}\right)\le0\)\(\Leftrightarrow\Sigma\frac{-30\left(a-1\right)^2}{5.25\left(2a+3\right)}\le0\) (đúng)
Ta có đpcm
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 ⇔ \(a=b=c=1\)
\(\left(a^2+3b^2\right)\left(1+3\right)\ge\left(a+3b\right)^2\Rightarrow\sqrt{a^2+3b^2}\ge\frac{a+3b}{2}\)
\(\Rightarrow P=\sum\frac{ab}{\sqrt{a^2+3b^2}}\le2\sum\frac{ab}{a+3b}=2\sum\frac{ab}{a+b+b+b}\)
\(\Rightarrow P\le\frac{1}{8}\sum ab\left(\frac{1}{a}+\frac{3}{b}\right)=\frac{1}{8}\sum\left(3a+b\right)=\frac{1}{2}\left(a+b+c\right)=\frac{3}{2}\)
"=" \(\Leftrightarrow a=b=c=1\)
\(1.\)\(a^3b^3\left(a^2-ab+b^2\right)\le\frac{\left(a+b\right)^8}{256}\)
\(\Leftrightarrow a^3b^3\left(a^2-ab+b^2\right)\left(a+b\right)\le\frac{\left(a+b\right)^9}{256}\)
\(\Leftrightarrow a^3b^3\left(a+b\right)^3\left(a^3+b^3\right)\le\frac{\left(a+b\right)^{12}}{256}\)
\(VT=ab\left(a+b\right).ab\left(a+b\right).ab\left(a+b\right).\left(a^3+b^3\right)\)
\(\le\left(\frac{ab\left(a+b\right)+ab\left(a+b\right)+ab\left(a+b\right)+\left(a^3+b^3\right)}{4}\right)^4\)
\(\le\frac{\left(a^3+3a^2b+3ab^2+b^3\right)^4}{256}\)
\(\le\frac{\left(a+b\right)^{12}}{256}\left(đpcm\right).\)
\(2.\) \(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\ge2\)
\(\Leftrightarrow\frac{1}{1+a}\ge1-\frac{1}{1+b}+1-\frac{1}{1+c}\)
\(\ge\frac{b}{1+b}+\frac{c}{1+c}\)
\(\ge2\sqrt{\frac{bc}{\left(1+b\right)\left(1+c\right)}}\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{1+b}\ge2\sqrt{\frac{ac}{\left(1+a\right)\left(1+c\right)}}\\\frac{1}{1+c}\ge2\sqrt{\frac{ab}{\left(1+a\right)\left(1+b\right)}}\end{cases}}\)
\(\Rightarrow\frac{1}{1+a}.\frac{1}{1+b}.\frac{1}{1+c}\ge8\sqrt{\frac{a^2b^2c^2}{\left(1+a\right)^2.\left(1+b\right)^2.\left(1+c\right)^2}}\)\(\frac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\ge\frac{8abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\)
\(\Leftrightarrow\) \(1\ge8abc\)
\(\Leftrightarrow\) \(abc\ge\frac{1}{8}\left(đpcm\right).\)
\(\frac{a}{3a+b+c}=\frac{a}{2a+a+b+c}\le\frac{1}{25}\left(\frac{4a}{2a}+\frac{9a}{a+b+c}\right)=\frac{2}{25}+\frac{9}{25}\left(\frac{a}{a+b+c}\right)\)
Tương tự: \(\frac{b}{a+3b+c}\le\frac{2}{25}+\frac{9}{25}\left(\frac{b}{a+b+c}\right)\) ; \(\frac{c}{a+b+3c}\le\frac{2}{25}+\frac{9}{25}\left(\frac{c}{a+b+c}\right)\)
Cộng vế với vế:
\(VT\le\frac{6}{25}+\frac{9}{25}\left(\frac{a+b+c}{a+b+c}\right)=\frac{3}{5}\)
Dấu "=" xảy ra khi \(a=b=c\)