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\(\Leftrightarrow\dfrac{a}{\sqrt{4b^2+bc+4c^2}}+\dfrac{b}{\sqrt{4c^2+ca+4a^2}}+\dfrac{c}{\sqrt{4a^2+ab+4b^2}}\ge1\)
Ta có:
\(\sum\left(\dfrac{a}{\sqrt{4b^2+bc+4c^2}}\right)^2\sum a\left(4b^2+bc+4c^2\right)\ge\left(a+b+c\right)^3\)
Nên ta chỉ cần chứng minh:
\(\dfrac{\left(a+b+c\right)^3}{a\left(4b^2+bc+4c^2\right)+b\left(4c^2+ac+4a^2\right)+c\left(4a^2+ab+4b^2\right)}\ge1\)
\(\Leftrightarrow\dfrac{\left(a+b+c\right)^3}{4a\left(b^2+c^2\right)+4b\left(c^2+a^2\right)+4c\left(a^2+b^2\right)+3abc}\ge1\)
\(\Leftrightarrow a^3+b^3+c^3+3abc\ge ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)\) (đúng theo Schur bậc 3)
\(\dfrac{1}{\left(a+b+a+c\right)^2}\le\dfrac{1}{4\left(a+b\right)\left(a+c\right)}=\dfrac{1}{4\left(a^2+ab+bc+ca\right)}\le\dfrac{1}{64}\left(\dfrac{1}{a^2}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\)
\(\le\dfrac{1}{64}\left(\dfrac{1}{a^2}+\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=\dfrac{1}{64}\left(\dfrac{2}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\)
Tương tự và cộng lại:
\(P\le\dfrac{1}{64}\left(\dfrac{4}{a^2}+\dfrac{4}{b^2}+\dfrac{4}{c^2}\right)=\dfrac{1}{16}.3=\dfrac{3}{16}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Áp dụng bđt: \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(1\right)\)
\(\dfrac{1}{2a+b+c}=\dfrac{1}{\left(a+b\right)+\left(a+c\right)}\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)\)
\(\Rightarrow P\le\dfrac{1}{16}\left[\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)^2+\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)^2+\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}\right)^2\right]\)\(\Rightarrow16P\le\dfrac{2}{\left(a+b\right)^2}+\dfrac{2}{\left(b+c\right)^2}+\dfrac{2}{\left(a+c\right)^2}+\dfrac{2}{\left(a+b\right)\left(b+c\right)}+\dfrac{2}{\left(a+b\right)\left(b+c\right)}+\dfrac{2}{\left(b+c\right)\left(c+a\right)}\)
Áp dụng: \(x^2+y^2+z^2\ge xy+yz+xz\left(2\right)\) với a+b=x,b+c=y,c+a=z
\(\Rightarrow16P\le\dfrac{4}{\left(a+b\right)^2}+\dfrac{4}{\left(b+c\right)^2}+\dfrac{4}{\left(c+a\right)^2}\)
Ta có: \(\dfrac{1}{\left(a+b\right)^2}\le4.16.\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2\)(do (1))
\(\Rightarrow16P\le\dfrac{1}{4}.16\left[\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2+\left(\dfrac{1}{b}+\dfrac{1}{c}\right)^2+\left(\dfrac{1}{c}+\dfrac{1}{a}\right)^2\right]=\dfrac{1}{4}\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}\right)\le\dfrac{1}{4}.4.\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=3\)(do(2) và \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=3\))
\(\Rightarrow P\le\dfrac{3}{16}\)
\(ĐTXR\Leftrightarrow a=b=c=1\)
Đề bài sai
Đề đúng: \(\dfrac{1}{\sqrt{a}+2\sqrt{b}+3}+\dfrac{1}{\sqrt{b}+2\sqrt{c}+3}+\dfrac{1}{\sqrt{c}+2\sqrt{a}+3}\le\dfrac{1}{2}\)
TK: Cho các số thực dương a, b, c thỏa mãn a + b+ c = 3. Chứng minh rằng: \(\sqrt{2a^2+\frac{7}{b^2}}+\sqrt{2b^2+\frac{7}{... - Hoc24
\(3\ge a^2+b^2+c^2\ge\dfrac{1}{3}\left(a+b+c\right)^2\Rightarrow a+b+c\le3\)
\(\Rightarrow\dfrac{a}{\sqrt[3]{3a+bc}}\le\dfrac{a}{\sqrt[3]{a\left(a+b+c\right)+bc}}=\sqrt[3]{2}.\sqrt[3]{\dfrac{a}{a+b}.\dfrac{a}{a+c}.\dfrac{a}{2}}\le\dfrac{\sqrt[3]{2}}{3}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{a}{2}\right)\)
Cộng vế và rút gọn:
\(E\le\dfrac{\sqrt[3]{2}}{3}\left(\dfrac{a}{a+b}+\dfrac{b}{a+b}+\dfrac{a}{a+c}+\dfrac{c}{a+c}+\dfrac{b}{b+c}+\dfrac{c}{b+c}+\dfrac{a+b+c}{2}\right)\)
\(E\le\dfrac{\sqrt[3]{2}}{3}\left(3+\dfrac{3}{2}\right)=\dfrac{3\sqrt[3]{2}}{2}\)
Thôi đang rảnh, giúp bạn bài này luôn vậy!!
Giải:
Ta có:
\(VT=\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)+\left(\dfrac{b^2}{b+c}+\dfrac{c^2}{c+a}+\dfrac{a^2}{a+b}\right)=A+B\)
\(A+3=\dfrac{1}{2}\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\left[\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right]\)
\(\ge\dfrac{1}{2}3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}3\sqrt[3]{\dfrac{1}{a+b}\dfrac{1}{b+c}\dfrac{1}{c+a}}=\dfrac{9}{2}\)
\(\Rightarrow A\ge\dfrac{3}{2}\)
\(1^2=\left(a+b+c\right)^2\le\left(\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{c+a}\right)\left(a+b+b+c+c+a\right)\)
\(\Leftrightarrow1\le B.2\Leftrightarrow B\ge\dfrac{1}{2}\)
Từ đó ta có: \(VT\ge\dfrac{3}{2}+\dfrac{1}{2}=2=VP\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
\(\dfrac{a+b^2}{b+c}+\dfrac{b+c^2}{c+a}+\dfrac{c+a^2}{a+b}\ge2\)
\(\Leftrightarrow\dfrac{a\left(a+b+c\right)+b^2}{b+c}+\dfrac{b\left(a+b+c\right)+c^2}{c+a}+\dfrac{c\left(a+b+c\right)+a^2}{a+b}\ge2\)
\(\Leftrightarrow\dfrac{a^2+ab+ac+b^2}{b+c}+\dfrac{ab+b^2+bc+c^2}{c+a}+\dfrac{ca+bc+c^2+a^2}{a+b}\ge2\)
\(\Leftrightarrow\dfrac{a^2+b^2+a\left(b+c\right)}{b+c}+\dfrac{b^2+c^2+b\left(c+a\right)}{c+a}+\dfrac{c^2+a^2+c\left(a+b\right)}{a+b}\ge2\)
\(\Leftrightarrow\dfrac{a^2+b^2}{b+c}+\dfrac{b^2+c^2}{c+a}+\dfrac{c^2+a^2}{a+b}+1\ge2\)
\(\Leftrightarrow\dfrac{a^2+b^2}{b+c}+\dfrac{b^2+c^2}{c+a}+\dfrac{c^2+a^2}{a+b}\ge1\)
\(\Leftrightarrow\dfrac{\sqrt{\left(a^2+b^2\right)^2}}{b+c}+\dfrac{\sqrt{\left(b^2+c^2\right)^2}}{c+a}+\dfrac{\sqrt{\left(c^2+a^2\right)^2}}{a+b}\ge1\)
Áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức
\(\Leftrightarrow\dfrac{\sqrt{\left(a^2+b^2\right)^2}}{b+c}+\dfrac{\sqrt{\left(b^2+c^2\right)^2}}{c+a}+\dfrac{\sqrt{\left(c^2+a^2\right)^2}}{a+b}\ge\dfrac{\left(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\right)^2}{2\left(a+b+c\right)}\)
\(\Leftrightarrow\dfrac{\sqrt{\left(a^2+b^2\right)^2}}{b+c}+\dfrac{\sqrt{\left(b^2+c^2\right)^2}}{c+a}+\dfrac{\sqrt{\left(c^2+a^2\right)^2}}{a+b}\ge\dfrac{\left(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\right)^2}{2}\)
Áp dụng bất đẳng thức Mincopski
\(\Rightarrow\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\ge\sqrt{2\left(a+b+c\right)^2}=\sqrt{2}\)
\(\Rightarrow\left(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\right)^2\ge2\)
\(\Rightarrow\dfrac{\left(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\right)^2}{2}\ge1\)
\(\Rightarrow\dfrac{\sqrt{\left(a^2+b^2\right)^2}}{b+c}+\dfrac{\sqrt{\left(b^2+c^2\right)^2}}{c+a}+\dfrac{\sqrt{\left(c^2+a^2\right)^2}}{a+b}\ge1\)
\(\Leftrightarrow\dfrac{a+b^2}{b+c}+\dfrac{b+c^2}{c+a}+\dfrac{c+a^2}{a+b}\ge2\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c=\dfrac{1}{3}\)